What is the integral of (8+4x)/(49+64x^2) ?

The integral of (8+4x)/(49+64x^2) is 1/7 arctan(8/7 x)+1/32 ln|49+64x^2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8+4x)/(49+64x^2)

\int \frac{8 + 4 x}{49 + 64 x ^{2}} d x

\frac{1}{7} arctan (\frac{8}{7} x) + \frac{1}{32} ln 49 + 64 x^{2} + C

Solution steps

\int \frac{8 + 4 x}{49 + 64 x ^{2}} d x

Expand

\frac{8 + 4 x}{49 + 64 x ^{2}} : \frac{8}{49 + 64 x ^{2}} + \frac{4 x}{49 + 64 x ^{2}}

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{8}{49 + 64 x ^{2}} d x + \int \frac{4 x}{49 + 64 x ^{2}} d x

\int \frac{8}{49 + 64 x ^{2}} d x = \frac{1}{7} arctan (\frac{8}{7} x)

\int \frac{4 x}{49 + 64 x ^{2}} d x = \frac{1}{32} ln 49 + 64 x^{2}

= \frac{1}{7} arctan (\frac{8}{7} x) + \frac{1}{32} ln 49 + 64 x^{2}

Add a constant to the solution

= \frac{1}{7} arctan (\frac{8}{7} x) + \frac{1}{32} ln 49 + 64 x^{2} + C

\int \frac{3 ^{x - 1}}{sec ( 3 ^{x} )} d x

\int sin (6 x) sin (8 x) d x

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

\int \frac{x}{( x ^{2} + 5 ) ^{\frac{7}{2}}} d x

\int x (5 - x^{2}) d x

What is the integral of (8+4x)/(49+64x^2) ?
The integral of (8+4x)/(49+64x^2) is 1/7 arctan(8/7 x)+1/32 ln|49+64x^2|+C