CE Board Exam Randomizer

Use $u=x$ and $dv=\cos(4x)\,dx$. Then $du=dx$ and $v=\frac{1}{4}\sin(4x)$.

$$\int x\cos(4x)\,dx=\frac{x}{4}\sin(4x)+\frac{1}{16}\cos(4x)$$

$$\left[\frac{x}{4}\sin(4x)+\frac{1}{16}\cos(4x)\right]_0^{\pi/4}=-\frac{1}{16}-\frac{1}{16}$$

$$\boxed{-\frac{1}{8}}$$

Since $x=t^2$, $dx=2t\,dt$. When $x=1$, $t=1$; when $x=4$, $t=2$.

$$xy\,dx=(t^2)(t^3)(2t\,dt)=2t^6\,dt$$

$$\int_1^4 xy\,dx=\int_1^2 2t^6\,dt=\left[\frac{2t^7}{7}\right]_1^2=\boxed{\frac{254}{7}=36.29}$$

Let $u=\tan^{-1}z$. Then $du=\frac{dz}{1+z^2}$.

$$\int_0^1 \frac{\tan^{-1}z}{1+z^2}\,dz=\int_0^{\pi/4}u\,du=\left[\frac{u^2}{2}\right]_0^{\pi/4}$$

$$\boxed{\frac{\pi^2}{32}}$$

Let $u=4-2x^2$, so $du=-4x\,dx$.

$$\int 5x(4-2x^2)^4\,dx=-\frac{5}{4}\int u^4\,du=-\frac{1}{4}(4-2x^2)^5+C$$

$$\boxed{-\frac{1}{4}(4-2x^2)^5+C}$$

Choose $u=\ln x$ and $dv=dx$. Then $du=\frac{1}{x}dx$ and $v=x$. Apply IBP: $\int u\,dv=uv-\int v\,du$.

$$\int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-\int 1\,dx$$

$$\boxed{x\ln x - x + C}$$

This is one of the most useful IBP results. Note the clever trick: treating $\ln x$ as "$u$" and the "invisible $dx$" as $dv$.

When IBP must be applied multiple times, build a table: differentiate the polynomial column to zero, integrate the exponential column each time. Alternate signs: +, −, +, −, +.

Multiply diagonally (sign × left × right), sum all:

$$\int x^3e^x\,dx=e^x(x^3-3x^2+6x-6)+C$$

$$\boxed{e^x(x^3-3x^2+6x-6)+C}$$

Apply IBP twice. First: $u=x^2$, $dv=\sin x\,dx$.

$$\int x^2\sin x\,dx=-x^2\cos x+2\int x\cos x\,dx$$

Second IBP: $u=x$, $dv=\cos x\,dx$:

$$\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x$$

Combining:

$$\boxed{-x^2\cos x+2x\sin x+2\cos x+C}$$

Apply IBP with $u=e^x$, $dv=\sin x\,dx$:

$$I=-e^x\cos x+\int e^x\cos x\,dx$$

Apply IBP again with $u=e^x$, $dv=\cos x\,dx$:

$$\int e^x\cos x\,dx=e^x\sin x-\int e^x\sin x\,dx=e^x\sin x-I$$

Substituting back: $I=-e^x\cos x+e^x\sin x-I$. Add $I$ to both sides:

$$2I=e^x(\sin x-\cos x)$$

$$\boxed{I=\frac{e^x(\sin x-\cos x)}{2}+C}$$

Use IBP: $u=\sec x$, $dv=\sec^2x\,dx$, so $du=\sec x\tan x\,dx$, $v=\tan x$.

$$\int\sec^3x\,dx=\sec x\tan x-\int\sec x\tan^2x\,dx$$

Replace $\tan^2x=\sec^2x-1$:

$$=\sec x\tan x-\int\sec^3x\,dx+\int\sec x\,dx$$

Let $I=\int\sec^3x\,dx$. Then $2I=\sec x\tan x+\ln|\sec x+\tan x|$:

$$\boxed{\int\sec^3x\,dx=\frac{\sec x\tan x+\ln|\sec x+\tan x|}{2}+C}$$

Step 1 — Complete the square: $x^2+6x+13=(x^2+6x+9)+4=(x+3)^2+4$.

$$\int\frac{dx}{(x+3)^2+4}$$

Step 2: Let $u=x+3$, $du=dx$. This matches the inverse-tangent form $\int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}(u/a)+C$ with $a=2$.

$$\boxed{\frac{1}{2}\tan^{-1}\!\left(\frac{x+3}{2}\right)+C}$$

Factor the denominator: $x^2-x-2=(x-2)(x+1)$.

Decompose: $\frac{3x+1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1}$.

Multiply through: $3x+1=A(x+1)+B(x-2)$. Set $x=2$: $7=3A\Rightarrow A=7/3$. Set $x=-1$: $-2=-3B\Rightarrow B=2/3$.

$$\int\frac{3x+1}{(x-2)(x+1)}\,dx=\frac{7}{3}\ln|x-2|+\frac{2}{3}\ln|x+1|+C$$

$$\boxed{\frac{1}{3}\ln\!\left|(x-2)^7(x+1)^2\right|+C}$$

Use the product-to-sum identity: $2\sin A\cos B=\sin(A-B)+\sin(A+B)$.

$$\sin(5x)\cos(3x)=\frac{1}{2}\left[\sin(2x)+\sin(8x)\right]$$

$$\int\sin(5x)\cos(3x)\,dx=\frac{1}{2}\int\left[\sin2x+\sin8x\right]dx$$

$$=\frac{1}{2}\left(-\frac{\cos2x}{2}-\frac{\cos8x}{8}\right)+C$$

$$\boxed{-\frac{\cos2x}{4}-\frac{\cos8x}{16}+C}$$

Odd power of sine — save one $\sin x$ for $du$, convert the rest using $\sin^2x=1-\cos^2x$:

$$\int\sin^2x\cos^2x\cdot\sin x\,dx=\int(1-\cos^2x)\cos^2x\sin x\,dx$$

Let $u=\cos x$, $du=-\sin x\,dx$:

$$=-\int(1-u^2)u^2\,du=-\int(u^2-u^4)\,du=-\frac{u^3}{3}+\frac{u^5}{5}+C$$

$$\boxed{-\frac{\cos^3x}{3}+\frac{\cos^5x}{5}+C}$$

The inside expression is $4-2x^2$, and its derivative is close to the outside factor $5x\,dx$.

Let $$u=4-2x^2.$$

Then $$du=-4x\,dx,$$ so $$x\,dx=-\frac{1}{4}du.$$

Substitute into the integral:

$$\int 5x(4-2x^2)^4dx=5\int u^4\left(-\frac{1}{4}du\right)=-\frac{5}{4}\int u^4du$$

$$=-\frac{5}{4}\left(\frac{u^5}{5}\right)+C=-\frac{u^5}{4}+C$$

Replace $u$ with $4-2x^2$:

$$\boxed{-\frac{1}{4}(4-2x^2)^5+C}$$

The denominator contains $1-2x^3$, whose derivative is $-6x^2$.

Let $$u=1-2x^3.$$

Then $$du=-6x^2\,dx,$$ so $$x^2\,dx=-\frac{1}{6}du.$$

Substitute:

$$\int \frac{5x^2}{(1-2x^3)^6}dx=5\int u^{-6}\left(-\frac{1}{6}du\right)=-\frac{5}{6}\int u^{-6}du$$

$$=-\frac{5}{6}\left(\frac{u^{-5}}{-5}\right)+C=\frac{1}{6}u^{-5}+C$$

$$\boxed{\frac{1}{6(1-2x^3)^5}+C}$$

The denominator is $y^2-4y+11$. Its derivative is $2y-4$.

The numerator can be factored as

$$5y-10=5(y-2)=\frac{5}{2}(2y-4).$$

Let $$u=y^2-4y+11.$$

Then $$du=(2y-4)dy.$$

Substitute:

$$\int \frac{5y-10}{y^2-4y+11}dy=\frac{5}{2}\int\frac{du}{u}$$

$$=\frac{5}{2}\ln|u|+C$$

Since $y^2-4y+11=(y-2)^2+7$ is always positive, the absolute value is optional here.

$$\boxed{\frac{5}{2}\ln(y^2-4y+11)+C}$$

Recognize the denominator as a perfect square:

$$4x^4-4x^2+1=(2x^2-1)^2.$$

Let $$u=2x^2-1.$$

Then $$du=4x\,dx,$$ so $$x\,dx=\frac{1}{4}du.$$

Substitute:

$$\int \frac{x\,dx}{(2x^2-1)^2}=\frac{1}{4}\int u^{-2}du$$

$$=\frac{1}{4}\left(-u^{-1}\right)+C=-\frac{1}{4u}+C$$

$$\boxed{-\frac{1}{4(2x^2-1)}+C}$$

The derivative of $\tan^{-1}z$ is $\frac{1}{1+z^2}$, which appears in the integrand.

Let $$u=\tan^{-1}z.$$

Then $$du=\frac{dz}{1+z^2}.$$

Change the limits: when $z=0$, $u=0$; when $z=1$, $u=\frac{\pi}{4}$.

$$\int_0^1\frac{\tan^{-1}z}{z^2+1}dz=\int_0^{\pi/4}u\,du$$

$$=\left[\frac{u^2}{2}\right]_0^{\pi/4}=\frac{1}{2}\left(\frac{\pi}{4}\right)^2$$

$$\boxed{\frac{\pi^2}{32}}$$

Let the expression inside the power be the new variable.

Let $$u=2z-3.$$

Then $$du=2\,dz,$$ so $$dz=\frac{1}{2}du.$$

Substitute:

$$\int(2z-3)^{-3/2}dz=\frac{1}{2}\int u^{-3/2}du$$

Use the power rule:

$$\frac{1}{2}\left(\frac{u^{-1/2}}{-1/2}\right)+C=-u^{-1/2}+C$$

$$\boxed{-\frac{1}{\sqrt{2z-3}}+C}$$

The inside expression is $1+4e^{3x}$. Its derivative is $12e^{3x}$.

Let $$u=1+4e^{3x}.$$

Then $$du=12e^{3x}dx,$$ so $$e^{3x}dx=\frac{1}{12}du.$$

Substitute:

$$\int(1+4e^{3x})^3e^{3x}dx=\frac{1}{12}\int u^3du$$

$$=\frac{1}{12}\left(\frac{u^4}{4}\right)+C=\frac{u^4}{48}+C$$

$$\boxed{\frac{(1+4e^{3x})^4}{48}+C}$$

Because the numerator has higher degree than the denominator, divide first.

$$\frac{2x^2-7x-5}{2x+1}=x-4-\frac{1}{2x+1}$$

Now integrate each part:

$$\int\left(x-4-\frac{1}{2x+1}\right)dx$$

$$=\frac{x^2}{2}-4x-\frac{1}{2}\ln|2x+1|+C$$

The factor $\frac{1}{2}$ appears because the derivative of $2x+1$ is $2$.

$$\boxed{\frac{x^2}{2}-4x-\frac{1}{2}\ln|2x+1|+C}$$

Rewrite the numerator in terms of the denominator. We want

$$w+1=A(5w+11)+B.$$

Matching coefficients gives $A=\frac{1}{5}$ and $B=-\frac{6}{5}$.

So

$$\frac{w+1}{5w+11}=\frac{1}{5}-\frac{6}{5(5w+11)}.$$

Integrate:

$$\int\left(\frac{1}{5}-\frac{6}{5(5w+11)}\right)dw=\frac{w}{5}-\frac{6}{5}\cdot\frac{1}{5}\ln|5w+11|+C$$

$$\boxed{\frac{w}{5}-\frac{6}{25}\ln|5w+11|+C}$$

Rewrite $x$ using the denominator $3x-4$.

$$x=\frac{1}{3}(3x-4)+\frac{4}{3}$$

Divide by $3x-4$:

$$\frac{x}{3x-4}=\frac{1}{3}+\frac{4}{3(3x-4)}$$

Integrate:

$$\int\left(\frac{1}{3}+\frac{4}{3(3x-4)}\right)dx=\frac{x}{3}+\frac{4}{3}\cdot\frac{1}{3}\ln|3x-4|+C$$

$$\boxed{\frac{x}{3}+\frac{4}{9}\ln|3x-4|+C}$$

Divide the numerator by $x^2+4$:

$$\frac{x^5-x^3+2x}{x^2+4}=x^3-5x+\frac{22x}{x^2+4}$$

Now integrate the polynomial terms directly:

$$\int x^3dx=\frac{x^4}{4},\qquad \int -5x\,dx=-\frac{5}{2}x^2$$

For the remaining term, let $u=x^2+4$, so $du=2x\,dx$:

$$\int\frac{22x}{x^2+4}dx=11\int\frac{du}{u}=11\ln|u|$$

Since $x^2+4$ is always positive:

$$\boxed{\frac{x^4}{4}-\frac{5}{2}x^2+11\ln(x^2+4)+C}$$

A useful trick is to split the numerator so it resembles the denominator:

$$e^{2x}-1=(e^{2x}+1)-2.$$

Then

$$\frac{e^{2x}-1}{e^{2x}+1}=1-\frac{2}{e^{2x}+1}.$$

A compact antiderivative is $\ln(e^{2x}+1)-x$. Check by differentiating:

$$\frac{d}{dx}\left[\ln(e^{2x}+1)-x\right]=\frac{2e^{2x}}{e^{2x}+1}-1$$

$$=\frac{2e^{2x}-(e^{2x}+1)}{e^{2x}+1}=\frac{e^{2x}-1}{e^{2x}+1}$$

Therefore,

$$\boxed{\ln(e^{2x}+1)-x+C}$$

The derivative of $\ln x$ is $\frac{1}{x}$, which appears in the integrand.

Let $$u=\ln x.$$

Then $$du=\frac{dx}{x}.$$

Substitute:

$$\int \frac{dx}{x\ln^2x}=\int u^{-2}du$$

Apply the power rule:

$$\int u^{-2}du=\frac{u^{-1}}{-1}=-u^{-1}$$

Return to $x$:

$$\boxed{-\frac{1}{\ln x}+C}$$

The inside expression is $1+\sin\beta$, and its derivative is $\cos\beta$.

Let $$u=1+\sin\beta.$$

Then $$du=\cos\beta\,d\beta.$$

Change the limits: when $\beta=0$, $u=1$; when $\beta=\frac{\pi}{2}$, $u=2$.

$$\int_0^{\pi/2}\frac{\cos\beta}{\sqrt{1+\sin\beta}}d\beta=\int_1^2u^{-1/2}du$$

$$=\left[2u^{1/2}\right]_1^2=2\sqrt{2}-2$$

$$\boxed{2(\sqrt{2}-1)}$$

Use the expression under the radical as $u$.

Let $$u=\cos2x-1.$$

Then $$du=-2\sin2x\,dx,$$ so $$\sin2x\,dx=-\frac{1}{2}du.$$

Substitute:

$$\int\frac{\sin2x}{\sqrt{\cos2x-1}}dx=-\frac{1}{2}\int u^{-1/2}du$$

$$=-\frac{1}{2}(2u^{1/2})+C=-\sqrt{u}+C$$

$$\boxed{-\sqrt{\cos2x-1}+C}$$

Domain reminder: $\cos2x-1\le0$, so this antiderivative is a formal real-radical expression only where the radicand is nonnegative. As written, the real-valued form is restricted.

The expression $\sqrt[3]{x}+5=x^{1/3}+5$ is the inside expression.

Let $$u=x^{1/3}+5.$$

Then $$du=\frac{1}{3}x^{-2/3}dx.$$

Since $x^{-2/3}dx=\frac{dx}{x^{2/3}}$, we have

$$\frac{dx}{x^{2/3}}=3du.$$

Substitute:

$$\int \frac{dx}{x^{2/3}(x^{1/3}+5)^{3/2}}=3\int u^{-3/2}du$$

$$=3\left(\frac{u^{-1/2}}{-1/2}\right)+C=-6u^{-1/2}+C$$

$$\boxed{-\frac{6}{\sqrt{\sqrt[3]{x}+5}}+C}$$

The derivative of $\tan(e^x+1)$ is $\sec^2(e^x+1)\cdot e^x$.

Let $$u=\tan(e^x+1).$$

Then $$du=e^x\sec^2(e^x+1)dx.$$

Substitute:

$$\int \frac{e^x\sec^2(e^x+1)}{\tan(e^x+1)}dx=\int\frac{du}{u}$$

$$=\ln|u|+C$$

$$\boxed{\ln|\tan(e^x+1)|+C}$$

The denominator is $3-e^{5x}$. Its derivative is $-5e^{5x}$.

Let $$u=3-e^{5x}.$$

Then $$du=-5e^{5x}dx,$$ so $$e^{5x}dx=-\frac{1}{5}du.$$

Substitute:

$$\int \frac{2e^{5x}}{3-e^{5x}}dx=2\int\frac{-\frac{1}{5}du}{u}$$

$$=-\frac{2}{5}\int\frac{du}{u}=-\frac{2}{5}\ln|u|+C$$

$$\boxed{-\frac{2}{5}\ln|3-e^{5x}|+C}$$

The derivative of $5x$ is $5$, so divide by $5$ when integrating $e^{5x}$.

$$\int 3e^{5x}dx=3\int e^{5x}dx=3\left(\frac{1}{5}e^{5x}\right)+C$$

$$\boxed{\frac{3}{5}e^{5x}+C}$$

The inside function is $\tan(4x)$, whose derivative is $4\sec^2(4x)$.

Let $$u=\tan(4x).$$

Then $$du=4\sec^2(4x)dx,$$ so $$\sec^2(4x)dx=\frac{1}{4}du.$$

$$\int 3^{\tan(4x)}\sec^2(4x)dx=\frac{1}{4}\int3^u\,du$$

Use $\int a^u du=\frac{a^u}{\ln a}+C$:

$$\boxed{\frac{3^{\tan(4x)}}{4\ln3}+C}$$

Rewrite the radical using exponents:

$$\sqrt[3]{e^{2y}}=e^{2y/3}.$$

So the integrand becomes

$$\frac{1}{e^{2y/3}}=e^{-2y/3}.$$

Integrate:

$$\int e^{-2y/3}dy=\frac{e^{-2y/3}}{-2/3}+C$$

$$\boxed{-\frac{3}{2}e^{-2y/3}+C}$$

Let the exponent be the new variable.

Let $$u=4-3x.$$

Then $$du=-3dx,$$ so $$dx=-\frac{1}{3}du.$$

$$\int e^{4-3x}dx=-\frac{1}{3}\int e^u du=-\frac{1}{3}e^u+C$$

$$\boxed{-\frac{1}{3}e^{4-3x}+C}$$

Let $$u=e^x.$$

Then $$du=e^x dx.$$

Since $e^{3x}dx=e^{2x}(e^x dx)=u^2du$, the integral becomes

$$\int\frac{u^2}{u-1}du.$$

Divide: $$\frac{u^2}{u-1}=u+1+\frac{1}{u-1}.$$

Integrate:

$$\int\left(u+1+\frac{1}{u-1}\right)du=\frac{u^2}{2}+u+\ln|u-1|+C$$

Return to $x$:

$$\boxed{\frac{1}{2}e^{2x}+e^x+\ln|e^x-1|+C}$$

The exponent is $\frac{1}{x}$. Its derivative is $-\frac{1}{x^2}=-x^{-2}$.

Let $$u=\frac{1}{x}.$$

Then $$du=-x^{-2}dx,$$ so $$x^{-2}dx=-du.$$

$$\int x^{-2}e^{1/x}dx=-\int e^u du=-e^u+C$$

$$\boxed{-e^{1/x}+C}$$

Convert $5^x$ to base $e$:

$$5^x=e^{x\ln5}.$$

Then

$$\frac{5^x}{e^{2x}}=\frac{e^{x\ln5}}{e^{2x}}=e^{(\ln5-2)x}.$$

Integrate using $\int e^{kx}dx=\frac{1}{k}e^{kx}+C$:

$$\int e^{(\ln5-2)x}dx=\frac{e^{(\ln5-2)x}}{\ln5-2}+C$$

$$\boxed{\frac{5^x}{(\ln5-2)e^{2x}}+C}$$

The expression inside the cube root is $4-e^{3x}$.

Let $$u=4-e^{3x}.$$

Then $$du=-3e^{3x}dx,$$ so $$e^{3x}dx=-\frac{1}{3}du.$$

Substitute:

$$\int e^{3x}(4-e^{3x})^{1/3}dx=-\frac{1}{3}\int u^{1/3}du$$

$$=-\frac{1}{3}\left(\frac{3}{4}u^{4/3}\right)+C=-\frac{1}{4}u^{4/3}+C$$

$$\boxed{-\frac{1}{4}(4-e^{3x})^{4/3}+C}$$

Divide each term by $e^{2x}$:

$$\frac{3-4e^{5x}}{e^{2x}}=3e^{-2x}-4e^{3x}.$$

Integrate term by term:

$$\int3e^{-2x}dx=-\frac{3}{2}e^{-2x}$$

$$\int-4e^{3x}dx=-\frac{4}{3}e^{3x}$$

$$\boxed{-\frac{3}{2}e^{-2x}-\frac{4}{3}e^{3x}+C}$$

Expand the square first:

$$(e^{2x}-e^{-3x})^2=e^{4x}-2e^{-x}+e^{-6x}.$$

Integrate each exponential term:

$$\int e^{4x}dx=\frac{1}{4}e^{4x}$$

$$\int -2e^{-x}dx=2e^{-x}$$

$$\int e^{-6x}dx=-\frac{1}{6}e^{-6x}$$

$$\boxed{\frac{1}{4}e^{4x}+2e^{-x}-\frac{1}{6}e^{-6x}+C}$$

The exponent is $\cot(x/3)$.

Let $$u=\cot(x/3).$$

Since $\frac{d}{dx}\cot(x/3)=-\frac{1}{3}\csc^2(x/3)$,

$$du=-\frac{1}{3}\frac{dx}{\sin^2(x/3)}.$$

So $$\frac{dx}{\sin^2(x/3)}=-3du.$$

Substitute:

$$\int \frac{5^{\cot(x/3)}}{\sin^2(x/3)}dx=-3\int5^u du$$

$$=-3\left(\frac{5^u}{\ln5}\right)+C$$

$$\boxed{-\frac{3}{\ln5}5^{\cot(x/3)}+C}$$

The exponent is $1-2x^2$, whose derivative is $-4x$.

Let $$u=1-2x^2.$$

Then $$du=-4x\,dx,$$ so $$x\,dx=-\frac{1}{4}du.$$

$$\int8^{1-2x^2}x\,dx=-\frac{1}{4}\int8^u du$$

$$=-\frac{1}{4}\cdot\frac{8^u}{\ln8}+C$$

$$\boxed{-\frac{8^{1-2x^2}}{4\ln8}+C}$$

Rewrite the cube root:

$$\sqrt[3]{5^{2y}}=5^{2y/3}.$$

Use $\int a^{ky}dy=\frac{a^{ky}}{k\ln a}+C$ with $a=5$ and $k=\frac{2}{3}$:

$$\int5^{2y/3}dy=\frac{5^{2y/3}}{(2/3)\ln5}=\frac{3}{2\ln5}5^{2y/3}.$$

Evaluate from $1$ to $3$:

$$\left[\frac{3}{2\ln5}5^{2y/3}\right]_1^3=\frac{3}{2\ln5}\left(5^2-5^{2/3}\right)$$

$$\boxed{\frac{3}{2\ln5}\left(25-5^{2/3}\right)}$$

Expand the square:

$$(4-e^{-z})^2=16-8e^{-z}+e^{-2z}.$$

Find an antiderivative:

$$\int(16-8e^{-z}+e^{-2z})dz=16z+8e^{-z}-\frac{1}{2}e^{-2z}.$$

Evaluate from $0$ to $3$:

$$\left[16z+8e^{-z}-\frac{1}{2}e^{-2z}\right]_0^3$$

$$=\left(48+8e^{-3}-\frac{1}{2}e^{-6}\right)-\left(8-\frac{1}{2}\right)$$

$$\boxed{\frac{81}{2}+8e^{-3}-\frac{1}{2}e^{-6}}$$

Integrate each term separately. Here $e$ is just a constant in the term $xe$.

$$\int e^{2x}dx=\frac{1}{2}e^{2x}$$

$$\int xe\,dx=e\frac{x^2}{2}$$

So

$$\int_0^2(e^{2x}+xe)dx=\left[\frac{1}{2}e^{2x}+\frac{e}{2}x^2\right]_0^2$$

$$=\left(\frac{e^4}{2}+2e\right)-\frac{1}{2}$$

$$\boxed{\frac{e^4-1}{2}+2e}$$

The denominator contains $5^{2x}-1$. Its derivative is proportional to $5^{2x}$.

Let $$u=5^{2x}-1.$$

Then $$du=2\ln5\cdot5^{2x}dx.$$

So $$5^{2x}dx=\frac{1}{2\ln5}du.$$

Substitute:

$$\int \frac{5^{2x}}{(5^{2x}-1)^3}dx=\frac{1}{2\ln5}\int u^{-3}du$$

$$=\frac{1}{2\ln5}\left(\frac{u^{-2}}{-2}\right)+C=-\frac{1}{4\ln5}u^{-2}+C$$

$$\boxed{-\frac{1}{4\ln5}\frac{1}{(5^{2x}-1)^2}+C}$$