# Problem in integration

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• December 14th 2010, 01:53 AM
kumaran5555
Problem in integration
$\int \frac{x^3}{(x^2+5)^2} dx
$

$
u=x^2
$

$
du = 2xdx.\frac{x^2}{x^2}
$

$
\frac{udu}{2}=x^3dx
$

$\frac{1}{2}\int \frac {udu}{(u+5)^2}$

$\frac{1}{2}[\int \frac {du}{u+5} - \int \frac{5du}{(u+5)^2}]$

$\frac{1}{2}[ln(u+5) +\frac{5}{(u+5)}]$

$\frac{1}{2}[ln(x^2+5) +\frac{5}{(x^2+5)}]$

But the answer is different.

Please help me finding which step is wrong.
• December 14th 2010, 02:05 AM
e^(i*pi)
According to Wolfram you forgot the constant of integration: $\frac{1}{2}[\ln(x^2+5) +\frac{5}{(x^2+5)}] + C$

The only real question I have is in the third line you've automatically assumed that $x \neq 0$ when multiplying by x^2/x^2
• December 14th 2010, 02:07 AM
Prove It
$\displaystyle \int{\frac{x^3}{(x^2 + 5)^2}\,dx} = \int{\frac{x^3}{x^4 + 10x^2 + 25}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{4x^3}{x^4 + 10x^2 + 25}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x - 20x}{x^4 + 10x^2 + 25}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x}{x^4 + 10x^2 + 25}\,dx} - \frac{1}{4}\int{\frac{20x}{x^4 + 10x^2 + 25}\,dx}$

$\displaystyle = \frac{1}{4}\int{\frac{4x^3 + 20x}{x^4 + 10x^2 + 25}\,dx} - \frac{5}{2}\int{\frac{2x}{(x^2 + 5)^2}\,dx}$.

Now for the first make the substitution $\displaystyle u = x^4 + 10x^2 + 25$ and for the second make the substitution $\displaystyle v= x^2 + 5$.
• December 14th 2010, 02:24 AM
kumaran5555
Quote:

Originally Posted by e^(i*pi)
According to Wolfram you forgot the constant of integration: $\frac{1}{2}[\ln(x^2+5) +\frac{5}{(x^2+5)}] + C$

The only real question I have is in the third line you've automatically assumed that $x \neq 0$ when multiplying by x^2/x^2

You mean to say that, the third step is wrong. What I have thought is, no matter what is the value of x, it can multipied and divided by a same value.

But i am clueless to understand your point.

Appreciate if you can help me.
• December 14th 2010, 02:35 AM
e^(i*pi)
Quote:

Originally Posted by kumaran5555
You mean to say that, the third step is wrong. What I have thought is, no matter what is the value of x, it can multipied and divided by a same value.

But i am clueless to understand your point.

Appreciate if you can help me.

By that logic $\frac{0}{0} = 1$ which is surely undefined as it's division by 0?
• December 14th 2010, 02:38 AM
kumaran5555
Thanks.

So, I should not proceed with the approach I have used.

Am I correct?
• December 14th 2010, 02:41 AM
Prove It
Quote:

Originally Posted by kumaran5555
Thanks.

So, I should not proceed with the approach I have used.

Am I correct?

You should proceed with the approach I posted...
• December 14th 2010, 02:44 AM
kumaran5555
Quote:

Originally Posted by Prove It
You should proceed with the approach I posted...

I have got your approach.

But my intention of this post is to understand what is wrong with my approach.

Thanks.
• December 14th 2010, 02:52 AM
kumaran5555
Quote:

Originally Posted by e^(i*pi)
By that logic $\frac{0}{0} = 1$ which is surely undefined as it's division by 0?

As you said 0/0 is undefined, multiplying x , with x/x is not going to cause any undefined value.

May be it's wrong multiplying a constant C with x/x and that may cause undefined results, when x=0.

But in mycase $x.\frac {x^2}{x^2}$
• December 14th 2010, 03:01 AM
e^(i*pi)
It will still be undefined when x=0. It would be fine if 0 wasn't in the original domain but because it is you're still potentially dividing by 0. This means that x=0 is a potential solution. In your equation let $f(x)= x \cdot \dfrac{x^2}{x^2}$ then $f(0) = 0 \cdot \dfrac{0}{0}$

You can do it if you exclude 0 from the domain
• December 14th 2010, 03:07 AM
kumaran5555
So, I am taking it as if any function which has zero in its domain, doing $f(x).\frac{x}{x}$ will result in undefined value. So it should not be done.
• December 14th 2010, 04:14 AM
e^(i*pi)
Yeah, that's it. Quite frustrating I admit