What is the integral of 1/(x^2)+8x-4 ?

The integral of 1/(x^2)+8x-4 is -1/x+4x^2-4x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2)+8x-4

\int \frac{1}{x ^{2}} + 8 x - 4 d x

- \frac{1}{x} + 4 x^{2} - 4 x + C

Solution steps

\int \frac{1}{x ^{2}} + 8 x - 4 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{2}} d x + \int 8 x d x - \int 4 d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int 8 x d x = 4 x^{2}

\int 4 d x = 4 x

= - \frac{1}{x} + 4 x^{2} - 4 x

Add a constant to the solution

= - \frac{1}{x} + 4 x^{2} - 4 x + C

\int - 2 x^{3} e^{4 x} d x

\int x^{2} \frac{1}{x ^{2} + 1} d x

\int \frac{1}{( 7 x - 6 ) ^{2}} d x

\int (x^{2} y + y + 2 x y) d y

\int (x^{2} - 3 x + 1)^{10} (2 x - 3) d x

What is the integral of 1/(x^2)+8x-4 ?
The integral of 1/(x^2)+8x-4 is -1/x+4x^2-4x+C