Intergration: Substitution - Question

**tttcomrader** · May 25th 2006, 04:03 PM

$ \int \frac{x}{1+x^4} \ dx $

I first started by setting $u = 1+x^4$ ,but then $du = 4x^3dx$ , doesn't work out with the $xdx$ that is left.

Any ideas?

By the way, I had finished my Spring Semester, with the help of this forum, I was able to pull As in my Calculus I and Linear Algebra. Just wanna thank you all!!!

**ThePerfectHacker** · May 25th 2006, 04:10 PM

Originally Posted by tttcomrader

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

I first started by setting $u = 1+x^4$ ,but then $du = 4x^3dx$ , doesn't work out with the $xdx$ that is left.

Let,
$u=1+x^2$ then, $\frac{du}{dx}=2x$
Therefore, express integral as,
$\frac{1}{2}\int \frac{1}{1+x^2}\cdot (2x) dx$
Substitute,
$\frac{1}{2}\int \frac{1}{u}\frac{du}{dx} dx=\frac{1}{2}\int \frac{1}{u} du=\frac{1}{2}\ln|u|+C$ Thus,
$\frac{1}{2}\ln |1+x^2|+C$
but, $1+x^2>0$ therefore,
$\frac{1}{2}\ln (1+x^2)+C$

**tttcomrader** · May 25th 2006, 07:05 PM

Oh, sorry, I made a mistake, the problem should be:

$ \int \frac{x}{1+x^4} \ dx $

Is to the 4th power instead of 2nd.

**ThePerfectHacker** · May 25th 2006, 07:17 PM

Originally Posted by tttcomrader

Oh, sorry, I made a mistake, the problem should be:

$ \int \frac{x}{1+x^4} \ dx $

Is to the 4th power instead of 2nd.

You have,
$\int \frac{x}{1+x^4}dx$
Let,
$u=x^2$ then, $\frac{du}{dx}=2x$
Express integrand as,
$\frac{1}{2}\int \frac{1}{1+(x^2)^2}\cdot (2x) dx$ and substitute,
$\frac{1}{2}\int \frac{1}{1+u^2}\frac{du}{dx}dx=\frac{1}{2}\int \frac{du}{1+u^2}=\frac{1}{2}\tan^{-1}(u)+C$
Substitute back,
$\frac{1}{2}\tan^{-1}(x^2)+C$

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