Question

Find the following integral : $\int x^2(1-\frac{1}{x^2}) \, dx$

Answer 1

Answer:

Step-by-step explanation:

Concept:

$1.\int[f(x)+g(x)]\:dx=\int{f(x)}\;dx+\int{g(x)}\;dx \\\\2.\int{x^n}\:dx=\frac{x^{n+1}}{n+1}+c$

Now,

$\int{x^2(1-\frac{1}{x^2})}\:dx \\\\=\int{[x^2-1]}\:dx \\\\=\int{x^2}\:dx-\int{1}\:dx \\\\=\frac{x^3}{3}-x+c$