2
$\begingroup$

I have to solve this problem using integration by parts. I am new to integration by parts and was hoping someone can help me.

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

Here is what I have so far:

$$\int udv = uv-\int vdu $$

$$u=x^2+2$$ Therefore, $$xdx=\frac{du}{2}$$ $$dv=x^3$$ Therefor, $$v=3x^2$$

$\endgroup$
2
  • 1
    $\begingroup$ so substitution isn't allowed? $\endgroup$
    – randomgirl
    Jul 13 '15 at 0:36
  • $\begingroup$ @randomgirl we have to use integration by parts $\endgroup$
    – Csci319
    Jul 13 '15 at 0:40
1
$\begingroup$

In integration by parts, you want $v$ to be something that is easy to integrate, $u$ is easy to differentiate, and $v\,du$ is easier to integrate than the original $u\,dv$.

In this case, the difficulty is integrating the fraction; and to make things more difficult, the fraction's denominator is squared. Here's one way to make that easier.

Let $dv=\frac{2x}{(x^2+2)^2}$. That is easy to integrate (can you see it?), and integrating it will reduce the power of the denominator. So then $u=\frac{x^2}2$ to get $u\,dv$ what we want.

Here is the rest:

We can see by inspection that the numerator of $dv$ is the derivative of the denominator. We therefore get $v=-\frac 1{x^2+2}$. Clearly $du=x\,dx$. So,

$$\begin{align} \int\frac{x^3}{(x^2+2)^2}\,dx &= uv-\int v\,du \\[2ex] &= \frac{x^2}2\cdot -\frac 1{x^2+2}-\int-\frac 1{x^2+2}\cdot x\,dx \\[2ex] &= -\frac 12\frac{x^2}{x^2+2}+\frac 12\int\frac{2x}{x^2+2}\,dx \\[2ex] &= -\frac 12\frac{x^2}{x^2+2}+\frac 12\ln|x^2+2|+C_1 \\[2ex] &= \frac{1}{x^2+2}-\frac 12+\frac 12\ln|x^2+2|+C_1 \\[2ex] &= \frac{1}{x^2+2}+\frac 12\ln|x^2+2|+C_2 \end{align}$$

Either the last line or the one two lines above it are good answers. Those two lines have different arbitrary constants, since the $-\frac 12$ disappeared between them.

Is that clear?

$\endgroup$
4
  • $\begingroup$ can you take a look and see if my answer is correct below? you have helped me in the past. $$\frac{-1}{2}\frac{x^2}{x^2+1}-(\frac{-1}{4}ln(|x^2+2|)$$ $\endgroup$
    – Csci319
    Jul 13 '15 at 14:53
  • $\begingroup$ yes please. I would like to know where I went wrong $\endgroup$
    – Csci319
    Jul 13 '15 at 15:24
  • $\begingroup$ @Csci319: Almost, one answer is $\frac{1}{x^2+2}+\frac 12\ln|x^2+2|+C$, another is $-\frac 12\frac{x^2}{x^2+2}+\frac 12\ln|x^2+2|+C$. Do you need me to show more steps in my answer? $\endgroup$ Jul 13 '15 at 15:28
  • $\begingroup$ please, it would help a lot $\endgroup$
    – Csci319
    Jul 13 '15 at 15:34
1
$\begingroup$

A couple of things with your work so far: When doing integration by parts, $u$ and $v$ have to be two functions which are multiplied by one another inside the integral. With your choice of $u$ and $v$, this is not the case. Second, to go from $dv$ to $v$, you should integrate, not differentiate. As a hint for how to proceed, you might try $$u=x^2$$ and $$dv=\frac{x}{(x^2+2)^2}dx.$$ Then $$du=2xdx$$ and $$v=-\frac 12\frac{1}{x^2+2}.$$ Can you take it from there?

$\endgroup$
8
  • $\begingroup$ so the integral which I need to solve will look like this: $$\int\frac{\frac{-1}{2}}{x^2+2}x^2$$? $\endgroup$
    – Csci319
    Jul 13 '15 at 0:52
  • $\begingroup$ Not quite. The integral you need to solve is $$\int v\,du=\frac{-1}{2}\int\frac{x}{x^2+2}dx.$$ $\endgroup$
    – Alex S
    Jul 13 '15 at 1:09
  • $\begingroup$ so then that would equal $$\frac{-1}{4}ln(|x^2+2|)+C$$? $\endgroup$
    – Csci319
    Jul 13 '15 at 2:09
  • $\begingroup$ Yes, but don't forget to add on $$uv=-\frac{1}{2}\frac{x^2}{x^2+1}.$$ $\endgroup$
    – Alex S
    Jul 13 '15 at 2:19
  • $\begingroup$ so $$\frac{-1}{2}\frac{x^2}{x^2+1}-(\frac{-1}{4}ln(|x^2+2|))+C$$? for the final answer $\endgroup$
    – Csci319
    Jul 13 '15 at 2:46
1
$\begingroup$

Let $dv=x/(x^2+2)^2$. and $u=x^2$. $v=\frac{-1/2}{x^2+2}$ and $du=2xdx$. You can now do the integral $\int vdu$ on your own, as it's just a logorithm.

$\endgroup$
2
  • $\begingroup$ so the integral which I need to solve will look like this: $$\int\frac{\frac{-1}{2}}{x^2+2}x^2$$? $\endgroup$
    – Csci319
    Jul 13 '15 at 0:52
  • $\begingroup$ that $x^2$ should be a $2x$ right? $\int udv = uv - \int vdu$. $\endgroup$
    – Alex R.
    Jul 13 '15 at 1:05
1
$\begingroup$

Hint:

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

Write $x^3asx^2x$

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

  • add and substract 2 in numerator

$\int\frac{[(x^2+2)-2]x}{(x^2+2)^2}dx$

  • separate it as two integrals

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

$=>\int\frac{x}{(x^2+2)}dx-\int\frac{2x}{(x^2+2)^2}dx$

formulae :(1) $\int\frac{f{'}(x)}{f(x)}dx=log|f(x)|$ formulae:(2)$int\frac{f{'}(x)}{f^{2}(x)}dx=-\frac{1}{f(x)}$

-take $f(x) = x^2+2$ for the first part of integral and before that multiply and divide with 2 for using formulae(1)

  • take $f(x) = x^2+2$for the second integral using formulae ( 2)

Then answer is

$=\frac{1}{2}log|x^2+2|+\frac{1}{x^2+2}$

$\endgroup$
1
  • $\begingroup$ Your answer is not what I would call a "hint". It is a walkthrough to solving the answer. Also, when you're writing equations, consider using \$\$math\$\$ instead of a single dollar sign. $\endgroup$ Jul 13 '15 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.