So i have this integral tending from 2(on top) to 1(on bottom ) of
$$\int_1^2x\ln(x^2 +3)x\,dx$$
If I turn it into $x^2\ln(x^2 +3)\,dx$, with integration by parts I get
$$\frac{16}{3}\,\ln(2) + \frac{49}{9}.$$
I was wondering, how is it possible that my TI nspire CAS does NOT recognize my integral? If I compute it, both with and without intervals, it gives me the same exact integral or a really weird equation with $\tan$ to the power of $-1.$
So I was wondering if any of you knew what could possibly be the problem. I tried typing it as the original integral, which is $x\ln(x^2 +3)x\,dx$, and also like this: $x^2\ln(x^2 +3)\,dx.$ Didn't get the answer above.
Any help would be really useful. :)
I am new here so mathjax is kind of a challenge to me... any help also on formatting the integral properly would be nice. :)
$\endgroup$ 141 Answer
$\begingroup$By parts
$$\begin{align}\int x^2\log(x^2+3)\,dx&=\frac{x^3}3\log(x^2+3)-\int\frac{2x^4}{3(x^2+3)}dx\\ &=\frac{x^3}3\log(x^2+3)-\frac23\int\left(x^2-3+\frac9{x^2+3}\right)dx\\ &=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-6\int\frac1{(\sqrt3t)^2+3}d(\sqrt3t)\\ &=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-\frac63\sqrt3\arctan t\\ &=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-2\sqrt3\arctan\frac x{\sqrt3} .\end{align}$$
$\endgroup$ 7
