3
$\begingroup$

How should I evaluate this definite integral? I am unable to figure out how to start.

$\int \tan^{-1} \left(1 + x + x^{2}\right) dx $

  • 1
    Integration by parts...2012-12-17

1 Answers 1

5

Proceed by integration by parts, we get \begin{align*} \int \arctan \left( 1 + x + x^2 \right) dx & = x \arctan \left(1+x+x^2 \right) - \int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx \end{align*} Now note that $\left(x^2+x+1 \right)^2+1 = \left(x^2+1 \right) \left(x^2+2x+2 \right)$ Hence, $\dfrac{x(2x+1)}{(x^2+x+1)^2+1} = \dfrac{x}{x^2+1} - \dfrac{x}{x^2 + 2x+2}$ Hence, $\int \dfrac{x(2x+1)}{(x^2+x+1)^2+1} dx = \int \dfrac{x dx}{x^2+1} - \int \dfrac{x dx}{x^2 + 2x+2}$ $\int \dfrac{x dx}{x^2+1} = \dfrac12 \log \left(1+x^2 \right)$ $\int \dfrac{x dx}{x^2 + 2x+2} = \int \dfrac{\left(x + 1 \right) dx}{\left( x+1 \right)^2+1} - \int \dfrac{dx}{\left( x+1 \right)^2+1} = \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$ Putting all this together, we get that $x \arctan \left( 1+x+x^2 \right) - \dfrac12 \log \left( 1+x^2\right) + \dfrac12 \log \left((x+1)^2+1\right) - \arctan(x+1)$ Rearranging gives us $x \arctan \left( 1+x+x^2 \right) - \arctan(x+1) + \dfrac12 \log \left(\dfrac{(x+1)^2+1}{x^2+1}\right) + \text{ constant}$

  • 0
    I edited the link. And of course that was just a joke.2012-12-17