Question

Integration of ∫x³+1/x³-x dx

Answer 1

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\int\limits (x^3+1)/(x^3-x) dx }} \end{gathered}$}`

We rewrite the integrand.

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{(x^3+1)/(x^3+x)=1+(1)/(x-1)-(1)/(x) }} \end{gathered}$}`

We integrate terms by terms:

The integral of the constants have this constant multiplied by the variable of integration.

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\int\limits 1 \ dx=x } } \end{gathered}$}`

That u = x -1. Then that du = dx and we put du.

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\int\limits(1)/(u) \ du }} \end{gathered}$}`

Integral 1/u is log(u). If you now substitute u further into:

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{log(x-1) }} \end{gathered}$}`

The integral of the product of a function by a constant is the constant times the integral of this function:

`\large\displaystyle\text{$\begin{gathered}\sf \boldsymbol{\sf{\int\limits\left(-(1)/(x)\right)dx=-\int\limits(1)/(x)dx }} \end{gathered}$}`

Integral 1/u is log(x). Therefore, the result is: -log(x).

The result is: x - log(x) + log(x - 1)