Question #a0c69

2 Answers
May 5, 2017

1/2(1+x^2)(ln(1+x^2)-1)+C12(1+x2)(ln(1+x2)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.

May 5, 2017

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)