could someone please check my work and see where i went wrong

Integral x^3/ (x^2+a^2) dx

Let u=x^2+a^2 where x^2=(u-a^2)

du=2x dx where dx=du/2x

Question

could someone please check my work and see where i went wrong

Integral x^3/ (x^2+a^2) dx 

Let u=x^2+a^2 where x^2=(u-a^2)

du=2x dx where dx=du/2x

anonymous · Answer

Integral x^3/ (x^2+a^2) dx = Integral x^3/ u du/2x = Integral x^2/u * du/2 
= 1/2 Integral (u-a^2)/u * du = 1/2 Integral (u/u - a^2/u) du 
= 1/2 Integral (1-a^2/u) du = 1/2 Integral 1du - 1/2 Integral a^2/u du
= 1/2x - 1/2a^2 Integral 1/u du = 1/2x - 1/2a^2 In(u) + c
= 1/2x - 1/2a^2 In(x^2+a^2) + c

ganeshie8 · Answer

check below highlighted line again :

= 1/2 Integral (u-a^2)/u * du = 1/2 Integral (u/u - a^2/u) du 
= 1/2 Integral (1-a^2/u) du = 1/2 Integral 1du - 1/2 Integral a^2/u du
`= 1/2x - 1/2a^2 Integral 1/u du = 1/2x - 1/2a^2 In(u) + c`
= 1/2x - 1/2a^2 In(x^2+a^2) + c

ganeshie8 · Answer

whats the integral of $1$ with respect to $u$ ?

anonymous · Answer

do you mean integral 1 du?

anonymous · Answer

@ganeshie8

ganeshie8 · Answer

exactly! $\large \int 1~ du = ?$

anonymous · Answer

it is x +c

ganeshie8 · Answer

who is x ?

ganeshie8 · Answer

we're integrating with respect to $u$ right ?
why bother about x here ?

anonymous · Answer

sorry u +c

ganeshie8 · Answer

Right, then do you see the mistake

anonymous · Answer

thank you so much i was a little bit confused

anonymous · Answer

ive been solving a lot of integrals i guess its time for bed

ganeshie8 · Answer

everything else is correct :) good job !!

ganeshie8 · Answer

wolfram is your best friend for checking integrals : http://www.wolframalpha.com/input/?i=Integral+x%5E3%2F+%28x%5E2%2Ba%5E2%29+dx+

anonymous · Answer

Thank you for that @ganeshie8  I really appreciate it

anonymous · Answer

i gave one small question when i plug in x^2+a^2 into u it comes up with a slightly different answer

anonymous · Answer

have*

anonymous · Answer

=1/2(x^2+a^2) - 1/2a^2 In(x^2+a^2) + c

ganeshie8 · Answer

somehow this is same as the answer given by wolfram :)
maybe differentiate ur final answer and see if you're getting back the integrand or not

ganeshie8 · Answer

did u get why they are both equal ? :)

ganeshie8 · Answer

notice that $\frac{1}{2}a^2$ is just a constant so you can pack it into the arbitrary constant $c$  :

\(\frac{1}{2}a^2 + c = C_2\

ganeshie8 · Answer

* $\frac{1}{2}a^2 + c = C_2 $

anonymous · Answer

ok so the a^2 has been taken out

ganeshie8 · Answer

yes because
$\frac{1}{2}a^2 + c$ is as arbitrary a constant as $c$ 

there is no need for two constant terms in the final answer

anonymous · Answer

if i had a million medals i would of given them to you :)

ganeshie8 · Answer

you can always pack the constant terms like this in the final answer

anonymous · Answer

I thank you again

ganeshie8 · Answer

np :)

anonymous · Answer

ill write it down as a note