What is the integral of 1/(64x^2-48x-7) ?

The integral of 1/(64x^2-48x-7) is -1/64 ln|8x+1|+1/64 ln|8x-7|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(64x^2-48x-7)

\int \frac{1}{64 x ^{2} - 48 x - 7} d x

- \frac{1}{64} ln ∣ 8 x + 1 ∣ + \frac{1}{64} ln ∣ 8 x - 7 ∣ + C

Solution steps

\int \frac{1}{64 x ^{2} - 48 x - 7} d x

Take the partial fraction of

\frac{1}{64 x ^{2} - 48 x - 7} : - \frac{1}{8 ( 8 x + 1 )} + \frac{1}{8 ( 8 x - 7 )}

= \int - \frac{1}{8 ( 8 x + 1 )} + \frac{1}{8 ( 8 x - 7 )} 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}{8 ( 8 x + 1 )} d x + \int \frac{1}{8 ( 8 x - 7 )} d x

\int \frac{1}{8 ( 8 x + 1 )} d x = \frac{1}{64} ln ∣ 8 x + 1 ∣

\int \frac{1}{8 ( 8 x - 7 )} d x = \frac{1}{64} ln ∣ 8 x - 7 ∣

= - \frac{1}{64} ln ∣ 8 x + 1 ∣ + \frac{1}{64} ln ∣ 8 x - 7 ∣

Add a constant to the solution

= - \frac{1}{64} ln ∣ 8 x + 1 ∣ + \frac{1}{64} ln ∣ 8 x - 7 ∣ + C

\int \frac{1}{25} d x

\int 2 x cos (2 x^{2}) d x

\int 7 cos (x) - \frac{8}{1 - x ^{2}} d x

\int x^{2} sin (x^{3} + 5) cos^{9} (x^{3} + 5) d x

\int x 1 + 16 x^{2} d x

What is the integral of 1/(64x^2-48x-7) ?
The integral of 1/(64x^2-48x-7) is -1/64 ln|8x+1|+1/64 ln|8x-7|+C