Answer to Question #164173 in Calculus for Mary Rose Delos Santos

Question

 (3x+1)dx / ((x+1)^12(3x-1)^12)

Answer 1

Solution:

Let $I=\int \dfrac{3x+1}{(x+1)^{12}(3x-1)^{12}}dx$

$=\int \dfrac{3x+1}{[(x+1)(3x-1)]^{12}}dx$

$=\int \dfrac{3x+1}{(3x^2+2x-1)^{12}}dx$ ...(i)

Now put $(3x^2+2x-1)=t$ ...(ii)

On differentiating w.r.t. $x$

$6x+2=\dfrac{dt}{dx}$

$\Rightarrow (3x+1)dx=\dfrac{dt}{2}$ ...(iii)

Putting (ii) and (iii) in (i), we get

$I=\int \dfrac{1}{(t)^{12}}\dfrac{dt}{2}$

$=\dfrac{1}{2}\int \dfrac{1}{t^{12}}dt$

$=\dfrac{1}{2}\int {t^{-12}}dt$

$=\dfrac{1}{2}(\dfrac{t^{-12+1}}{{-12+1}})+C$

$=\dfrac{1}{2}(\dfrac{t^{-11}}{{-11}})+C$

$=\dfrac{-1}{{22t^{11}}}+C$

$=\dfrac{-1}{{22(3x^2+2x-1)^{11}}}+C$ [Using (ii)]

Thus, our final answer is $\dfrac{-1}{{22(3x^2+2x-1)^{11}}}+C$