Integral

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May 24th 2006, 10:30 PM
AfterShock

Integral

I recall doing this integral during the academic year, and I remember getting it right after working on it for quite some time. Out of curiosity, is there any way to calculate it quickly, or is it just tedious integration by parts multiple times?

Integral(x^2*arctan(x))
May 24th 2006, 10:49 PM
earboth

Quote:

Originally Posted by AfterShock

I recall doing this integral during the academic year, and I remember getting it right after working on it for quite some time. Out of curiosity, is there any way to calculate it quickly, or is it just tedious integration by parts multiple times?
Integral(x^2*arctan(x))

Hello,

actually this problem is worse than you suggested:

1. step: Partial integration. Then you have to split the next integrand by synthetic division into 2 summands.

2. step: The remainder of the division is integrated by substitution.

3. The final result is:

$\int{x^2 \cdot \arctan(x)}dx={1\over 3} \cdot x^3 \cdot \arctan(x) + {1\over 6} \cdot \ln(x^2+1)-{1\over 6} \cdot x^2$

Greetings

EB
May 28th 2006, 05:05 AM
Soroban

Hello, AfterShock!

Did you try integration by parts?
. . It does not require repeated applications.

Here's what earboth suggested . . . in baby-steps.

Quote:

$\int x^2\arctan(x)\,dx$

Let: $u = \arctan(x)\qquad dv = x^2\,dx$

Then: $du = \frac{dx}{1 + x^2}\quad v = \frac{1}{3}x^3$

And we have: . $\frac{1}{3}x^3\arctan(x) - \frac{1}{3}\int\frac{x^3}{1+x^2}\,dx$

Use long division on that fraction: . $\frac{1}{3}x^3\arctan(x) - \frac{1}{3}\int\left(x - \frac{x}{1 + x^2}\right)\,dx$

And we have: . $\frac{1}{3}x^3\arctan(x) - \frac{1}{3}\int x\,dx + \frac{1}{3}\int\frac{x}{1 + x^2}\,dx$

And the answer is what earboth promised . . .