Integral of $\int x \cdot \sqrt{x+x^2}$

Question

Integrate the following

$$\int x \cdot \sqrt{x+x^2}dx$$

I used $x=x+\frac{1}{2}-\frac{1}{2}$ and then took $x+x^2=u$ and solved but it turned out to be very lengthy solution. Could someone suggest a easier approach?

Answer 1

hint: $x^2+ x = \left(x+\frac{1}{2}\right)^2 - \dfrac{1}{4}\implies u = x+\dfrac{1}{2}$, and followed by $v = \dfrac{\sec \theta}{2}$. This should reduce the integral to a sum of $2$ familiar integrals that you can look up in calc book easily or do it yourself with a simple substitution.

Answer 2

Hint:

$$\int x\sqrt{x+x^2}dx=\frac12\int(2x+1)\sqrt{x+x^2}dx-\frac12\int\sqrt{x+x^2}\,dx$$ where the first integral should be obvious ($f'\sqrt f$).

Then by completing the square

$$\int\sqrt{x+x^2}dx=\frac12\int\sqrt{(2x+1)^2-1}\,dx$$

and with $2x+1=\cosh u$,

$$\int\sqrt{(2x+1)^2-1}\,dx=\frac12\int \sinh^2u\,du.$$

The latter is easily solved using the identity

$$\sinh^2u=\frac{\cosh2u-1}2.$$

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Or, by direct application of the change of variable,

$$\int x\sqrt{x+x^2}dx=\frac14\int(\cosh u-1)\sinh^2u\,du=\frac1{12}\sinh^3u-\frac14\sinh2u+\frac u2.$$