Question

Find the following integral : $\int \frac{x^3+5x^2-4}{x^2} \, dx$

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,

$\int{[\frac{x^3+5x^2-4}{x^2}]}\:dx\\\\=\int[\frac{x^3}{x^2}+5\frac{x^2}{x^2}-4\frac{1}{x^2}]\:dx\\\\=\int[x+5-4x^{-2}]\:dx\\\\=\int{x}\:dx+5\int{dx}-4\int{x^{-2}}\:dx\\\\=\frac{x^2}{2}+5x-4(-2)x^{-3}+c\\\\=\frac{x^2}{2}+5x+8x^{-3}+c$