Question

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

Answer 1

`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