Question

How do you evaluate the integral `int x^2arctanx`?

Answer 1

The answer is `=x^3/3arctanx-color(blue)(1/6(-ln(1+x^2)+(1+x^2)))+C`

Explanation:

`sin^2x+cos^2x=1`, `=>`, `tan^2x+1=sec^2x`

If `y=arctanx`

`tany=x`

`(tany)'=x'`

`sec^2ydy/dx=1`, `=>`, `dy/dx=1/sec^2y=1/(1+x^2)`
We use integration by parts

`intuv'dx=uv-intu'vdx`

`u=arctanx`, `=>`, `u'=1/(1+x^2)`

`v'=x^2`, `=>`, `v=x^3/3`

Therefore,

`intx^2arctanxdx=x^3/3arctanx-color(blue)(1/3int(x^3dx)/(1+x^2))`

Let, u=1+x^2`,`=>`,`du=2xdx#

`x^2=u-1`

so,

`color(blue)(1/3int(x^3dx)/(1+x^2))=color(blue)(1/6*int(u-1)/udu)`

`=color(blue)(1/6(int-(du)/u+intdu))`

`=color(blue)(1/6(-lnu+u))`

`=color(blue)(1/6(-ln(1+x^2)+(1+x^2)))`

Finally, we have

`intx^2arctanxdx=x^3/3arctanx-color(blue)(1/6(-ln(1+x^2)+(1+x^2)))+C`

Answer 2

`x^3/3arc tanx-x^2/6+1/6ln(x^2+1)+C`.

Explanation:

Let `I=intx^2arc tanxdx`

We will use the following Rule of Integration by Parts (IBP) :

`IBP : intuvdx=uintvdx-int[(du)/dxintvdx]dx.`

We take `u=arc tanx :. (du)/dx=1/(x^2+1)`, and,

`v=x^2 :. intvdx=x^3/3`

Hence, `I=x^3/3arc tanx-1/3intx^3/(x^2+1)dx=x^3/3arc tanx-1/3J,`

where, `J=intx^3/(x^2+1)dx=int(x^3+x-x)/(x^2+1)dx`

`=int{x(x^2+1)}/(x^2+1)dx-1/2int(2x)/(x^2+1)dx`

`=intxdx-1/2int{d/dx(x^2+1)}/(x^2+1)dx`

`:. J=x^2/2-1/2ln(x^2+1)`

Altogether, `I=x^3/3arc tanx-1/3{x^2/2-1/2ln(x^2+1)}, or,`

`I=x^3/3arc tanx-x^2/6+1/6ln(x^2+1)+C`.

The Later Integral of J has been directly obtained using the

`" Result : "int (f'(x))/f(x)dx=ln|f(x)|+c`, which can easily be

proved by the substitution `f(x)=t`.

Enjoy Maths.!