Question

Question #a0c69

Answer 1

`1/2(1+x^2)(ln(1+x^2)-1)+C`

Explanation:

`int\ xln(1+x^2)\ dx`

Substitute `u=1+x^2` and `du=2x\ dx`.

`int\ (xln(u))/(2x)\ du=1/2int\ ln(u)\ du`

Use integration by parts: `int\ f\ dg=fg-int\ g\ df` with `f=ln(u),df=1/u\ du,dg=du,g=u`.

`1/2int\ ln(u)\ du=1/2(u ln(u)-int\ u*1/u\ du)=1/2(u ln(u)-int\ du)=1/2(u ln(u)-u+C)=1/2u(ln(u)-1)+C`.

Substitute `u=1+x^2` back.

`1/2(1+x^2)(ln(1+x^2)-1)+C`.

Answer 2

`intxln (1+x^2)dx=1/2 (1+x^2)(ln (1+x^2)-1)`

Explanation:

We know a formulae that
`intlntdt= t (lnt-1)`
Now let
`I=` `intxln (1+x^2)dx`
Substitute `1+x^2=t`
`=>2xdx=dt`
`:. I=1/2intlntdt`
`:.I= 1/2 t (lnt-1)`
`:.I=1/2 (1+x^2)(ln (1+x^2)-1)`