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

1 Answer
Feb 21, 2016

x\ln (x^2)-2x+\frac{x^2}{2}+C

Explanation:

\int \ln (x^2)+xdx

Applying sum rule:
=\int f(x)\pm g\(x)dx=\int f(x)dx\pm \int g(x)dx

=\int \ln (x^2)dx+\int \xdx.... (i)

=\int \ln (x^2)dx

Applying integration by parts:

=\int \uv'=uv-\int u'v

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

=\ln (x^2)x-\int \frac{2}{x}xd

=x\ln (x^2)-\int 2dx
(int 2dx= 2x)

=x\ln (x^2)-2x

Also,
=\int xdx=\frac{x^2}{2}
(Applying power rule, \int x^adx=\frac{x^{a+1}}{a+1})

Now substituting the value in equation (i),

=x\ln (x^2)-2x+\frac{x^2}{2}

=x\ln (x^2)-2x+\frac{x^2}{2} + C