Question

Solve

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

Answer 1

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

$\int {\left(\frac{x+1-2}{x+1}\right)}^4$

$\int {(1-\frac{2}{x+1})}^{4}dx$

$\int {(1-\frac{2}{x+1})}^2{(1-\frac{2}{x+1})}^{2}dx$

$\int (1+\frac{4}{{(x+1)}^2}-\frac{4}{x+1})(1+\frac{4}{{(x+1)}^2}-\frac{4}{x+1})dx$

$\int 1+\frac{4}{{(x+1)}^2}-\frac{4}{(x+1)}+\frac{4}{{(x+1)}^2}+\frac{16}{{(x+1)}^4}-\frac{16}{{(x+1)}^3}-\frac{4}{(x+1)}-\frac{16}{{(x+1)}^3}+\frac{16}{{(x+1)}^2}$dx

$\int 1+\frac{8+16}{{(x+1)}^2}-\frac{8}{x+1}+\frac{16}{{(x+1)}^4}-\frac{32}{{(x+1)}^3}dx$

$\int (1+24{(x+1)}^{-2}-8{(x+1)}^{-1}+16{(x+1)}^{-4}-32{(x+1)}^{-3})dx=x+\frac{24}{(x+1)(-2+1)}-8\log (x+1)-\frac{16}{3{(x+1)}^3}+\frac{32}{2{(x+1)}^3}$

$=x-\frac{24}{x+1}-8\log (x+1)-\frac{16}{3{(x+1)}^3}+\frac{16}{{(x+1)}^2}+c.$

Hence, the answer is $x-\frac{24}{x+1}-8\log (x+1)-\frac{16}{3{(x+1)}^3}+\frac{16}{{(x+1)}^2}+c.$