How To Integrate These?

1. Dec 1, 2014

basty

1. The problem statement, all variables and given/known data

FIRST
$\int x^2 \ln x \ dx$

SECOND
$\int \frac{(x+1)^2}{x}dx$

THIRD
$\int \frac{x-2}{x+3}dx$

2. Relevant equations

N/A

3. The attempt at a solution
I tried using the integration by part method but it doesn't work.

For the FIRST problem:
Setting $u=x^2$ or $u=\ln x$ is not solving the problem.

For the SECOND problem:
Setting $u=x+1$ or $u=x$ is not solving the problem.

For the THIRD problem:
Setting $u=x-2$ or $u=x+3$ is not solving the problem.

Last edited: Dec 1, 2014
2. Dec 1, 2014

Orodruin

Staff Emeritus
You need to provide more details than this on your attempted solution. Show us what you did and what you got.

3. Dec 1, 2014

basty

For the FIRST problem:
Setting $u=x^2$ or $u=\ln x$ is not solving the problem.

For the SECOND problem:
Setting $u=x+1$ or $u=x$ is not solving the problem.

For the THIRD problem:
Setting $u=x-2$ or $u=x+3$ is not solving the problem.

4. Dec 1, 2014

ehild

What is u? show how did you proceed in the first case.
How do you do the integration by parts?

5. Dec 1, 2014

basty

Give me some hints.

6. Dec 1, 2014

pasmith

Differentiating $\ln x$ and integrating $x^2$ is the correct method. What did you get when you tried it?

The remaining two problems do not require integration by parts. They can be done by direct integration after some algebraic manipulation of the integrand.

7. Dec 1, 2014

basty

Thank you

This is the FIRST problem's solution done by me after getting a hint from you:
let $u=\ln x$ then $\frac{du}{dx}=\frac{1}{x}$ or $du=\frac{1}{x}dx$
let $dv=x^2$ then $v=\frac{1}{3}x^3$

So $\int x^2 \ln x \ dx = u.v-\int v.du$
$=\ln x . \frac{1}{3}x^3-\int\frac{1}{3}x^3(\frac{1}{x}dx)$
$=\ln x . \frac{1}{3}x^3-\frac{1}{3}\int\frac{x^3}{x}dx$
$=\ln x . \frac{1}{3}x^3-\frac{1}{3}\int x^2dx$
$=\ln x . \frac{1}{3}x^3-\frac{1}{3}[\frac{1}{3}x^3+c]$
$=\ln x . \frac{1}{3}x^3-\frac{1}{9}x^3+c$
CMIIW.

How to do the algebraic manipulation? I don't understand. Give some hints again for the SECOND and THIRD problem.

8. Dec 1, 2014

ehild

expand the square in the second problem.

9. Dec 1, 2014

basty

You're right!

$\int\frac{(x+1)^2}{x}dx$
$=\int\frac{x^2+2x+1}{x}dx$
$=\int\frac{x^2}{x}dx+\int\frac{2x}{x}dx+\int\frac{1}{x}dx$
$=\int x \ dx+2\int dx+\int\frac{1}{x}dx$
$=\frac{1}{2}x^2+2x+\ln x+c$

Is it correct?

One more problem to solve.

Give me a hint.

10. Dec 1, 2014

ehild

Write the numerator as the sum (x+3) -5.

You have the integral
$\int \frac{(x+3)-5}{x+3}dx=\int 1-\frac{5}{x+3}dx$