UC Berkeley Integration Bee 2025 Qualifying Exam

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Source: UC Berkeley Integration Bee 2025

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456 posts

#1 h Apr 27, 2025, 1:47 AM

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Good luck and have fun!!!

1. $\int_{-\sqrt[3]{2}}^{1}(x^3+x^6)(x^2+2x^5)dx$
2. $\int \frac{e^x}{1+e^{2x}}+\frac{x}{1+x^2}dx$
3. $\int_0^\pi x^4\cos(x)dx$
4. $\int_{0}^{\frac{\pi}{2}}\sin^2(x)\cos^2(x)dx$
5. $\int_1^2\frac{\sqrt{x^2+2\sqrt{x^2-1}}}{\sqrt{x^2-1}}dx$
6. $\int_2^4\frac{(2-x)}{(x-1)(x-5)}dx$
7. $\int_0^\infty \frac{dx}{x^2-2x+2}$
8. $\int (e^x+\ln(x))\left(e^x+\frac{1}{x}\right)dx$
9. $\int_{-\infty}^{\infty}\frac{\ln(x^2+1)}{x^2}dx$
10. $\int_0^1 \frac{dx}{\sqrt{1+\sqrt{x}}-\sqrt{1-\sqrt{x}}}$
11. $\int_0^1 2^{\ln(x)}dx$
12. $\int \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] \cdot \frac{\ln(x)}{e^x} dx$
13. $\int_1^\infty \ln(x)-\frac{1}{2}\ln(x^2+1)dx$
14. $\int_0^1 2x\sqrt{x-x^2}dx$
15. $\int_0^1 \ln^3(x)dx$
16. $\int_0^\pi\frac{\sqrt{\csc(x)-\sin(x)}}{\sqrt{\sin(x)}+1}dx$
17. $\int_0^6\frac{x}{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}dx$
18. $\int_0^\infty \frac{\sin(x)\cos(2x)\cos^2(x)}{x}dx$
19. $\int_0^{\infty}\frac{dx}{1+x+x^2+x^3}$
20. $\int_{-\infty}^{\infty}\frac{(x^5-x^3)}{(x^2-1)^4+x^4}dx$

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#2 h Apr 27, 2025, 3:55 AM

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This post has been edited 10 times. Last edited by aidan0626, Apr 27, 2025, 6:20 AM

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techb

#3 h Apr 27, 2025, 5:29 AM

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5) $\int_1^2\frac{\sqrt{x^2+2\sqrt{x^2-1}}}{\sqrt{x^2-1}}dx$

$I = \int_1^2\frac{\sqrt{x^2-1+2\sqrt{x^2-1}+1}}{\sqrt{x^2-1}}dx = \int_1^2\frac{\sqrt{(\sqrt{x^2-1})^2+2\sqrt{x^2-1}+1}}{\sqrt{x^2-1}}dx = \int_1^2\frac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}}dx = \int_1^2 \bigg(1+\frac{1}{\sqrt{x^2-1}}\bigg)dx$
We will use the fact $\int \frac{1}{\sqrt{x^2-1}}dx = \ln|x+\sqrt{x^2-1}| + C$ (It can easily be derived using elementary methods.)

$I = \int_1^2 \bigg(1+\frac{1}{\sqrt{x^2-1}}\bigg)dx = x + \ln|x+\sqrt{x^2-1}| \bigg|_1^2 = \boxed{1+\ln(2+\sqrt{3})}$

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techb

#4 h Apr 27, 2025, 5:41 AM

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6) $\int_2^4\frac{(2-x)}{(x-1)(x-5)}dx$
Using king's method,
$I = \int_2^4\frac{(x-4)}{(x-1)(x-5)}dx$ $2I = \int_2^4\frac{-2}{(x-1)(x-5)}dx$ $I = \int_2^4\frac{-1}{(1-x)(5-x)}dx = \frac{1}{4}\int_2^4\bigg(\frac{1}{5-x} - \frac{1}{1-x} \bigg)dx = \frac{1}{4}\ln\bigg|\frac{1-x}{5-x}\bigg|\bigg|_2^4 = \boxed{\frac{\ln3}{2}}$

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techb

#5 h Apr 27, 2025, 6:03 AM

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8) $\int (e^x+\ln(x))\left(e^x+\frac{1}{x}\right)dx$
Use u-substitution of $u= e^x+\ln(x)$ , $du = \bigg(e^x+\frac{1}{x}\bigg)dx$ .

$I = \int (e^x+\ln(x))\left(e^x+\frac{1}{x}\right)dx = \int u du = \frac{u^2}{2} + C = \boxed{\frac{(e^x+\ln(x))^2}{2} + C}$

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techb

#6 h Apr 27, 2025, 6:06 AM • 1 Y

Y by aidan0626

12) $\int \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] \cdot \frac{\ln(x)}{e^x} dx$
Use u-substitution of $u = \frac{e^x}{\ln(x)}$ , $du = \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] dx$

$I = \int \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] \cdot \frac{\ln(x)}{e^x} dx = \int \frac{1}{u} du = \boxed{\ln\bigg(\frac{e^x}{\ln(x)}\bigg)+C}$

This post has been edited 1 time. Last edited by techb, Apr 27, 2025, 6:07 AM

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#7 h Apr 27, 2025, 6:12 AM

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12) $\int \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] \cdot \frac{\ln(x)}{e^x} dx$
Use u-substitution of $u = \frac{e^x}{\ln(x)}$ , $du = \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] dx$

$I = \int \frac{d}{dx}\left [ \frac{e^x}{\ln(x)} \right] \cdot \frac{\ln(x)}{e^x} dx = \int \frac{1}{u} du = \boxed{\ln\bigg(\frac{e^x}{\ln(x)}\bigg)+C}$

oh wait that's a smarter way than how i did it oops

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techb

#8 h Apr 27, 2025, 6:13 AM

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10) $\int_0^1 \frac{dx}{\sqrt{1+\sqrt{x}}-\sqrt{1-\sqrt{x}}}$
$I = \int_0^1 \frac{dx}{\sqrt{1+\sqrt{x}}-\sqrt{1-\sqrt{x}}} = \int_0^1 \frac{\big(\sqrt{1+\sqrt{x}}+\sqrt{1-\sqrt{x}}\big)dx}{2\sqrt{x}}$
Use u-substitution of $u=\sqrt{x}$ , $du = \frac{1}{2\sqrt{x}}dx$

$I = \int_0^1 \big(\sqrt{1+u}+\sqrt{1-u}\big)du = \frac{2}{3} ((1+u)^{3/2} - (1-u)^{3/2}) \bigg|_0^1 = \boxed{\frac{4\sqrt{2}}{3}}$

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#9 h Apr 27, 2025, 6:37 AM

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1. 0
2. arctan(e^x)+ 1/2 (ln(1+x^2))+C

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#10 h Apr 27, 2025, 11:40 AM

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3.

$I=\int x^{4}\cos x\ dx$ , four times integration by parts:
$I=x^{4}\sin x+4x^{3}\cos x-12x^{2}\sin x-24x\cos x+24\sin x +C$ .
Result: $-4\pi^{3}+24\pi$ .

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#11 h Apr 27, 2025, 11:51 AM

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9.

$I=\int_{-\infty}^{\infty}\frac{\ln(x^2+1)}{x^2}\ dx$ , integration by parts:
$I=\left[-\frac{\ln(x^{2}+1)}{x}\right]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{1}{x} \cdot \frac{2x}{x^{2}+1}\ dx$ .
$I=0+2\left[\arctan x\right]_{-\infty}^{\infty} =2\pi$ .

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#12 h Apr 27, 2025, 12:12 PM

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13.

$I=\int_{1}^{\infty} \left[\ln x-\frac{1}{2}\ln(x^{2}+1)\right]\ dx$ .
Each term, integration by parts:
$I=\left[x\ln x-x\right]_{1}^{\infty}-\frac{1}{2}\left(\left[x\ln(x^{2}+1)\right]_{1}^{\infty}-\int_{1}^{\infty} x \cdot \frac{2x}{x^{2}+1}\ dx\right)$ .
$I=\left[x\ln x-x\right]_{1}^{\infty}-\left(\frac{1}{2}\left[x\ln(x^{2}+1)\right]_{1}^{\infty}-\int_{1}^{\infty} \frac{x^{2}}{x^{2}+1}\ dx\right)$ .
$I=\left[x\ln x-x\right]_{1}^{\infty}-\left[\frac{1}{2}x\ln(x^{2}+1)-x +\arctan x\right]_{1}^{\infty}$ .
$I=\left[x\ln x-x\ln \sqrt{x^{2}+1}-\arctan x\right]_{1}^{\infty}$ .
$I=\left[x\ln \frac{x}{\sqrt{x^{2}+1}}-\arctan x\right]_{1}^{\infty}=\ln \frac{1}{\sqrt{2}}-\frac{\pi}{4}$ .

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3ch03s

#13 h Apr 27, 2025, 4:53 PM

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14. Do $u=2x+1$ and parity to get
$\int_0^1 2x\sqrt{x-x^2}dx=\dfrac{1}{4}\int_{-1}^1 (u-1)\sqrt{1-u^2}du=\dfrac{1}{4}\int_{-1}^1 u\sqrt{1-u^2}du+\dfrac{1}{4}\int_{-1}^1\sqrt{1-u^2}du=\dfrac{\pi}{8}$
17.Let $y=\sqrt{x+\sqrt{x+...}}$ , then $x=y^2-y$ , but as $y\geq 0$ and $x \in [0,6]$ by a simple inequation we obtain $y \in\{0\}\cup [1.3]$ . but we neglected the point 0. Hence:

$\int_0^6\frac{x}{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}dx=\int_1^3 \dfrac{y^2-y}{y}(2y-1)dy=\int_1^3 (y-1)(2y-1)dy=\int_0^2 u(2u+1)du=\int_0^2 2u^2+u=\dfrac{16}{3}+2=\dfrac{22}{3}$
19. (V2)
$\int_0^{\infty}\frac{dx}{1+x+x^2+x^3}=\int_0^{\infty}\frac{\frac 12(1+x^2)+\frac{1}{2}(1-x^2)}{(1+x)(1+x^2)}dx=\dfrac 12 \int_0^\infty \dfrac{1}{1+x}dx+\dfrac{1-x}{1+x^2}dx$ $=\dfrac 12(\ln(1+x)+\arctan(x)-\dfrac 12(1+x^2)|_0^{\infty}$ Now: $\ln(1+x)-\dfrac 12 \ln(1+x^2)=\ln(1+1/x)-\dfrac 12\ln (1+1/x^2)$ and hence the limit at $\infty$ is the same as the limit at 0 from the right, which is 0. Therefore
$=\dfrac 12(\ln(1+x)+\arctan(x)-\dfrac 12(1+x^2)|_0^{\infty}=\dfrac{\pi}{4}$
20. The integrand is an odd function and $\int_0^{\infty}\frac{(x^5-x^3)}{(x^2-1)^4+x^4}dx$ exists by comparison with $1/x^3$ (denominator is always >0) hence the integral is 0. Eff' you glasser.

Regards
Claudio.

This post has been edited 2 times. Last edited by 3ch03s, Apr 27, 2025, 5:15 PM
Reason: Need some justification.

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#14 h Apr 27, 2025, 7:12 PM

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16.

$I=\int_{0}^{\pi} \frac{\sqrt{\csc x-\sin x}}{\sqrt{\sin x}+1}\ dx=2\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\csc x-\sin x}}{\sqrt{\sin x}+1}\ dx$ .
$I=2\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(\sqrt{\sin x}+1)\sqrt{\sin x}}\ dx$ .

Substitution: $x=\sin t$ , then
$I=2\int_{0}^{1} \frac{1}{(\sqrt{t}+1)\sqrt{t}}\ dt$ .

Substitution= $t=z^{2}$ , then
$I=2\int_{0}^{1} \frac{2z}{(z+1)z}\ dt=4\left[\ln(z+1)\right]_{0}^{1}=4\ln 2$ .

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#15 h Apr 27, 2025, 7:38 PM

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20.

$I=\int_{-\infty}^{\infty} \frac{(x^5-x^3)}{(x^2-1)^4+x^4}\ dx$ .

Substitution: $x=-t$ , then
$I=\int_{-\infty}^{\infty} \frac{(-t^5+t^3)}{(t^2-1)^4+t^4}\ dt=-I$ , so $I=0$ .

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#16 h Apr 28, 2025, 1:00 PM

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18.

It's well known that
$\int_0^\infty\frac{\sin(x)}x\,\mathrm dx = \frac\pi2,$ and by substitution $x\to ax$ for $a>0$ we get
$\int_0^\infty\frac{\sin(ax)}x\,\mathrm dx = \frac\pi2$ as well. Then
$\int_0^\infty\frac{\sin(x)\cos(2x)\cos^2(x)}x\,\mathrm dx = \frac12\int_0^\infty\frac{\sin(2x)\cos(2x)\cos(x)}x\,\mathrm dx$ $= \frac14\int_0^\infty\frac{\sin(4x)\cos(x)}x\,\mathrm dx = \frac18\int_0^\infty\frac{\sin(5x)+\sin(3x)}x\,\mathrm dx = \frac18\left(\frac\pi2+\frac\pi2\right) = \frac\pi8.$
19. (alternative solution)
We can do the keyhole contour for
$f(z) = \frac{\log_{(0,2\pi)}(z)}{1+z+z^2+z^3}.$ This fcn has poles at $-1$ , $i$ and $-i$ with residues $\pi i/2$ , $\pi(1-i)/8$ and $-3\pi(1+i)/8$ respectively, and the integrals of $f(z)$ around $0$ and $\infty$ vanish in the limit, so we get
$\int_0^\infty\frac{\ln(x)}{1+x+x^2+x^3}\,\mathrm dx-\int_0^\infty\frac{\ln(x)+2\pi i}{1+x+x^2+x^3}\,\mathrm dx =2\pi i\left(\frac{\pi i}2+\frac{\pi(1-i)}8-\frac{3\pi(1+i)}8\right),$ hence
$\int_0^\infty\frac{\mathrm dx}{1+x+x^2+x^3} = -\left(\frac{\pi i}2+\frac{\pi(1-i)}8-\frac{3\pi(1+i)}8\right) = \frac\pi4.$

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#17 h Apr 29, 2025, 1:31 AM

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The actual solution for 19.

$=\int_0^\infty \frac{dx}{1+x+x^2+x^3} = \int_0^\infty \frac{dx}{(1+x)(1+x^2)}$ Let $x=\tan\theta$ and then Queen's rule by $u=\frac{\pi}{2}-x$ .

$=\int_0^{\frac{\pi}{2}}\frac{d\theta}{1+\tan\theta} = \boxed{\frac{\pi}{4}}$

This post has been edited 1 time. Last edited by Silver08, Apr 29, 2025, 5:29 AM

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3ch03s

#18 h Apr 29, 2025, 3:15 AM

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Silver08 wrote:

The actual solution for 18.

$=\int_0^\infty \frac{dx}{1+x+x^2+x^3} = \int_0^\infty \frac{dx}{(1+x)(1+x^2)}$ Let $x=\tan\theta$ and then Queen's rule by $u=\frac{\pi}{2}-x$ .

$=\int_0^{\frac{\pi}{2}}\frac{d\theta}{1+\tan\theta} = \boxed{\frac{\pi}{4}}$

18 or 19?

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#19 h Apr 29, 2025, 5:29 AM

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3ch03s wrote:

18 or 19?

oh crap 19, thank you

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