# Thread: need help with integration by parts

1. ## need help with integration by parts

Evaluate the following:

• $\displaystyle \int\frac{(ln t)^2}{t}\, dt$
• $\displaystyle \int\ln(x^2 + 1)\, dx$
• $\displaystyle \int\sin t\, ln(cos t)\, dt$
• $\displaystyle \int\frac{cot^-1\sqrt{z}} {\sqrt{z}}\,dz$
• $\displaystyle \int\sqrt{x}\ ln x\,dx$

2. $\displaystyle u=(ln t)^2$
$\displaystyle dv=\dfrac{1}{t}$
$\displaystyle du=\dfrac{2(ln t)}{t}$
$\displaystyle v=ln |t|$

$\displaystyle \int\frac{(ln t)^2}{t}\, dt = ln|t|(ln t)^2-\int\frac{2(ln t)^2}{t}\, dt$
$\displaystyle 3\int\frac{(ln t)^2}{t}\, dt = {ln|t|(ln t)^2}$
$\displaystyle \int\frac{(ln t)^2}{t}\, dt = \dfrac{(ln x)^3}{3}$

get the idea?

3. Originally Posted by cazimi
Evaluate the following:
$\displaystyle \int\ln(x^2 + 1)\, dx$
Define $\displaystyle f(x)=x\ln(x^2+1)\implies f'(x)=\ln(x^2+1)+\frac{2x^2}{x^2+1}.$

Now $\displaystyle \frac{{2x^2 }} {{x^2 + 1}} = 2\left( {1 - \frac{1} {{x^2 + 1}}} \right).$

Integrate $\displaystyle f'(x),$

$\displaystyle f(x)+k=\int\ln(x^2+1)\,dx+2(x-\arctan x).$

And finally $\displaystyle \int\ln(x^2+1)\,dx=x\ln(x^2+1)-2(x-\arctan x)+k.$

4. Let $\displaystyle u=ln(x^2+1)$ and $\displaystyle v = x$

Thus,
$\displaystyle du=\dfrac{2x}{x^2+1}dx$ and $\displaystyle dv = dx$

$\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-\int\dfrac{2x^2}{x^2+1}\, dx$
$\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2\int\dfrac{x^2+1-1}{x^2+1}\, dx$
$\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2\int1-\dfrac{1}{1+x^2}\, dx$
$\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2(x-arctan\:x)$
Therefore $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2x+arctan\:x+C$

5. The first one does not require integration by parts, just define $\displaystyle u=\ln t$ and the rest follows.

6. Originally Posted by cazimi
$\displaystyle \int\sqrt{x}\ ln x\,dx$
Consider the function $\displaystyle f(x)=x^{3/2}\ln x\implies f'(x)=\frac32\sqrt x\ln x+\sqrt x.$

Now apply some of make-up,

$\displaystyle \frac23f'(x)=\sqrt x\ln x+\frac23\sqrt x.$

Integrate, $\displaystyle \frac23f(x)+k=\int\sqrt x\ln x\,dx+\frac49x^{3/2}.$

Back substitute

$\displaystyle \int\sqrt x\ln x\,dx=\frac23x^{3/2}\ln x-\frac49x^{3/2}+k.$

7. Let $\displaystyle u=ln(cos\:x)$ and $\displaystyle dv=sin\:dx$
Thus, $\displaystyle du=\dfrac{sin\:x}{cos\:x}dx$ and $\displaystyle v=-cos\:x$

$\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:xln(cos\:x)-\int\dfrac{-sin\:x\:cos\:x}{cos\:x}dx$
$\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:xln(cos\:x)+\int\sin\:x\:dx$

Therefore $\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:x[ln(cos\:x)+1] + C$

8. Thank you!

9. ## integration by parts

find the following genreal anti-derivatives using integration by parts formula

\int\frac{(x+ 2}{e^3x}\, dx

10. Originally Posted by Sundevils
find the following genreal anti-derivatives using integration by parts formula

\int\frac{(x+ 2}{e^3x}\, dx

Please do not double post. See rule #1 here.

-Dan