Question

Solve $\int \frac{x^4+4x^3+11x^2+12x+8}{(x^2+2x+3{)}^2(x+1)}dx$

Answer 1

Let $I=\int \frac{x^4+4x^3+11x^2+12x+8}{(x^2+2x+3{)}^2(x+1)}dx$
Now $(x^2+2x+3{)}^2=x^4+4x^2+9+4x^3+12x+6x^2$
$=x^4+4x^3+10x^2+12x+9=N^r-(x^2-1)\therefore N^r=(x^2+2x+3{)}^2+(x^2-1)$
$I=\int [\frac{1}{x+1}+\frac{x-1}{(x^2+2x+3{)}^2}]dx=log(x+1)+\int \frac{l(2x+2)+m}{(x^2+2x+3{)}^2}dx$
where $2l=1,2l+m=-1\Rightarrow l=\frac{1}{2},m=-2$
$\therefore I=log(x+1)-\frac{l}{(x^2+2x+3)}+m\int \frac{1}{\left[\right(x+1{)}^2+{\sqrt{2}}^2{]}^2}dx$
$=log(x+1)-\frac{1}{2(x^2+2x+3)}-2l_1$
For $l_1$ put $x+1=\sqrt{2}\tan \theta$
$\therefore I_1=\int \frac{\sqrt{2}{\sec }^2\theta d\theta }{4{\sec }^4\theta }=\frac{1}{4\sqrt{2}}\int 2{\cos }^2\theta d\theta =\frac{1}{4\sqrt{2}}\int (1+\cos 2\theta )d\theta$
$=\frac{1}{4\sqrt{2}}[\theta +\frac{\sin 2\theta }{2}]=\frac{1}{4\sqrt{2}}[{\tan }^{-1}\frac{x+1}{\sqrt{2}}+\frac{(x+1)\sqrt{2}}{(x^2+2x+3)}]$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\frac{x+1}{\sqrt{2}}+\frac{1}{2}\frac{x+1}{x^2+2x+3}$
Hence $I=log(x+1)-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x+1}{\sqrt{2}}+\frac{x+1}{x^2+2x+3}$