What is int ln(x)^2+xdx?

Feb 21, 2016

$x \setminus \ln \left({x}^{2}\right) - 2 x + \setminus \frac{{x}^{2}}{2} + C$

Explanation:

$\setminus \int \setminus \ln \left({x}^{2}\right) + x \mathrm{dx}$

Applying sum rule:
=$\setminus \int f \left(x\right) \setminus \pm g \setminus \left(x\right) \mathrm{dx} = \setminus \int f \left(x\right) \mathrm{dx} \setminus \pm \setminus \int g \left(x\right) \mathrm{dx}$

=$\setminus \int \setminus \ln \left({x}^{2}\right) \mathrm{dx} + \setminus \int \setminus x \mathrm{dx}$.... (i)

=$\setminus \int \setminus \ln \left({x}^{2}\right) \mathrm{dx}$

Applying integration by parts:

=$\setminus \int \setminus u v ' = u v - \setminus \int u ' v$

=$u = \setminus \ln \left({x}^{2}\right) , \setminus u ' = \setminus \frac{2}{x} , v ' = 1 , v = x$

=$\setminus \ln \left({x}^{2}\right) x - \setminus \int \setminus \frac{2}{x} x d$

=$x \setminus \ln \left({x}^{2}\right) - \setminus \int 2 \mathrm{dx}$
($\int 2 \mathrm{dx}$= 2x)

=$x \setminus \ln \left({x}^{2}\right) - 2 x$

Also,
=$\setminus \int x \mathrm{dx} = \setminus \frac{{x}^{2}}{2}$
(Applying power rule, $\setminus \int {x}^{a} \mathrm{dx} = \setminus \frac{{x}^{a + 1}}{a + 1}$)

Now substituting the value in equation (i),

=$x \setminus \ln \left({x}^{2}\right) - 2 x + \setminus \frac{{x}^{2}}{2}$

=$x \setminus \ln \left({x}^{2}\right) - 2 x + \setminus \frac{{x}^{2}}{2}$ + C