# `int_1^2 (ln(x))^2/x^3 dx` Evaluate the integral

`int_1^2(ln(x))^2/x^3dx`

If f(x) and g(x) are differentiable function, then

`intf(x)g'(x)=f(x)g(x)-intf'(x)g(x)dx`

If we rewrite f(x)=u and g'(x)=v, them

`intuvdx=uintvdx-int(u'intvdx)dx`

Using the above method of integration by parts,

Let's first evaluate the indefinite integral,

`int(ln(x))^2/x^3dx=(ln(x))^2int1/x^3dx-int(d/dx(ln(x))^2int(1/x^3)dx)dx`

`=(ln(x))^2(x^(-3+1)/(-3+1))-int(2ln(x)(1/x)(x^(-3+1)/(-3+1)))dx`

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

`=-1/(2x^2)(ln(x))^2+int(ln(x)/x^3)dx`

again applying integration by parts

`=-1/(2x^2)(ln(x))^2+ln(x)*int(1/x^3)dx-int(d/dx(ln(x)int(1/x^3)dx)dx`

`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-3+1)/(-3+1))-int(1/x)((x^-3+1)/(-3+1))dx`

`=-1/(2x^2)(ln(x))^2+ln(x)*(x^(-2)/-2)-int(1/(-2x^3))dx`

`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2int(1/x^3)dx`

`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)+1/2(x^(-3+1)/(-3+1))`

`=-1/(2x^2)(ln(x))^2-ln(x)/(2x^2)-1/(4x^2)`

add a constant C to the solution,

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

Now let's evaluate definite integral,

`int_1^2(ln(x))^2/x^3dx=[-1/(2x^2)((ln(x))^2+ln(x)+1/2)]_1^2`

`=[-1/(2*2^2)((ln(2))^2+ln(2)+1/2)]-[-1/(2*1^2)((ln(1))^2+ln(1)+1/2)]`

`=[-1/8(ln(2))^2-ln(2)/8-1/16+1/4]`

`=3/16-(ln(2))^2/8-ln(2)/8`

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