(x^8-x^2)dx(x^12+3x^6+1) ^-1(x^3+1x^3)

Question

$\int \frac{({x}^{8}-{x}^{2})dx}{({x}^{12}+3{x}^{6}+1){	an }^{-1}({x}^{3}+\frac{1}{{x}^{3}})}$

Accepted Answer

$I=\int \frac{{x}^{8}-{x}^{2}}{({x}^{12}+3{x}^{6}+1){	an }^{-1}({x}^{3}+\frac{1}{{x}^{3}})}dx$

$\begin{gathered} {	an }^{-1}({x}^{3}+\frac{1}{{x}^{3}})=t \ \Rightarrow \frac{1}{1+{({x}^{3}+\frac{1}{{x}^{3}})}^{2}}\cdot (3{x}^{2}-\frac{3}{{x}^{4}})dx=dt \ \Rightarrow \frac{{x}^{6}}{{x}^{12}+3{x}^{6}+1}\cdot \frac{3{x}^{6}-3}{{x}^{4}}dx=dt \ I=\frac{1}{3}\int \frac{dt}{t}=\frac{1}{3}\ln |t|+C \ I=\frac{1}{3}\ln |{	an }^{-1}({x}^{3}+\frac{1}{{x}^{3}})|+C \ I=\ln {|{	an }^{-1}({x}^{3}+\frac{1}{{x}^{3}})|}^{1/3}+C \end{gathered}$

Hence option (1) is correct.

Solutions · 1