Question

Evaluate: $\int {\left(\frac{x-1}{x+1}\right)}^{4}dx$

Answer 1

Let $I=\int {\left(\frac{x-1}{x+1}\right)}^{4}dx$

$\frac{(x-1{)}^4}{(x+1{)}^4}=\frac{A}{x+1}+\frac{B}{(x+1{)}^2}+\frac{C}{(x+1{)}^3}+\frac{D}{(x+1{)}^4}$

$(x-1{)}^4=A(x+1{)}^3+B(x+1{)}^2+C(x+1)+D$

When $x=-1,$ $(-2{)}^4=D\Rightarrow D=16$

when $x=0,$ $1=A+B+C+D$

when $x=1,$ $0=(2{)}^{3}A+(2{)}^{2}B+2C+D$

$\Rightarrow 0=8A+4B+2C+D$

When $x=2,$ $1=27A+9B+3C+D$

$A+B+C=-15$

$4A+2B+C=-8$

$9A+3B+C=-5$

$\Rightarrow 3A+B=7$

Also, $4A+B=5$

$\Rightarrow A=-2,B=13,C=-26$

Let $x+1=t$

$dx=dt$

$\Rightarrow I=-2\int \frac{dt}{t}+13\int \frac{dt}{t^2}-26\int \frac{dt}{t^3}+16\int \frac{dt}{t^4}$

$\Rightarrow I=-2\log t-\frac{13}{t}+\frac{13}{t^2}-\frac{16}{3t^3}+c$

$\Rightarrow I=-2\log (x+1)-\frac{13}{x+1}+\frac{13}{(x+1{)}^2}-\frac{16}{3(x+1{)}^3}+c$