What is the integral of (3x^2+x)/((8x^3+4x^2)^2) ?

The integral of (3x^2+x)/((8x^3+4x^2)^2) is 1/(16x)-1/(32x^2)-1/(8(2x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2+x)/((8x^3+4x^2)^2)

\int \frac{3 x ^{2} + x}{( 8 x ^{3} + 4 x ^{2} ) ^{2}} d x

\frac{1}{16 x} - \frac{1}{32 x ^{2}} - \frac{1}{8 ( 2 x + 1 )} + C

Solution steps

\int \frac{3 x ^{2} + x}{( 8 x ^{3} + 4 x ^{2} ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + x}{( 8 x ^{3} + 4 x ^{2} ) ^{2}} : - \frac{1}{16 x ^{2}} + \frac{1}{16 x ^{3}} + \frac{1}{4 ( 2 x + 1 ) ^{2}}

= \int - \frac{1}{16 x ^{2}} + \frac{1}{16 x ^{3}} + \frac{1}{4 ( 2 x + 1 ) ^{2}} 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}{16 x ^{2}} d x + \int \frac{1}{16 x ^{3}} d x + \int \frac{1}{4 ( 2 x + 1 ) ^{2}} d x

\int \frac{1}{16 x ^{2}} d x = - \frac{1}{16 x}

\int \frac{1}{16 x ^{3}} d x = - \frac{1}{32 x ^{2}}

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

= - (- \frac{1}{16 x}) - \frac{1}{32 x ^{2}} - \frac{1}{8 ( 2 x + 1 )}

Simplify

= \frac{1}{16 x} - \frac{1}{32 x ^{2}} - \frac{1}{8 ( 2 x + 1 )}

Add a constant to the solution

= \frac{1}{16 x} - \frac{1}{32 x ^{2}} - \frac{1}{8 ( 2 x + 1 )} + C

\int (2 x - 6) d x

\int 4 x y + 3 x^{2} d x

\int x^{2} (x^{3} + 4)^{2} d x

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

\int \frac{2 x + 3}{3 x + 2} d x

What is the integral of (3x^2+x)/((8x^3+4x^2)^2) ?
The integral of (3x^2+x)/((8x^3+4x^2)^2) is 1/(16x)-1/(32x^2)-1/(8(2x+1))+C