Question

integrate of (2x^3/1+x^8)dx,X€R​

Answer 1

Answer:

hope it helps youiuuuu

Answer 2

Given:

We have given an equation of integeration $\int\limits {\frac{2x^3}{1+x^8} } \, dx$.

To Find:

We have to integrate the given equation?

Step-by-step explanation:

• We have equation which is written below

        $\int\limits {\frac{2x^3}{1+x^8} } \, dx$

• For integrating the above equation first of all we will let the value

        $\textrm{Let} ,x^4=t$

• Now do the differentiation on the both side of the above equation we get

       $4x^3dx=dt$

• Now take 4 on the other side from above equation.

        $x^3dx=\frac{dt}{4}$

• We will put the value of x to the power cube dx in the given integration equation we get.

         $\int\limits {\frac{2x^3}{1+x^8} } \, dx =\int\limits {\frac{2dt}{4(1+t^2)} } \, dx$

• Now simplify the above equation.

       $\Rightarrow \int\limits {\frac{dt}{2(1+t^2)} } \, dx$

• Now take the constant term outside.

      $\Rightarrow \frac{1}{2} \int\limits {\frac{dt}{(1+t^2)} } \, dx$

• We will use the formula of integration in above equation which is

       $\int\limits {\frac{1}{1+x^2} } \, dx =tan^-^1x+C$

• By using above formula we get.

       $\Rightarrow \frac{1}{2} (tan^-^1t)+C$

• Now put the value of t in the above equation.

       $\Rightarrow \frac{1}{2} (tan^-^1x^4)+C$

Thus, integration of the above equation is $\Righarrow \frac{1}{2} (tan^-^1x^4)+C$.