What is the integral of 6/((x^2+1)(x^2+4)) ?

The integral of 6/((x^2+1)(x^2+4)) is 2arctan(x)-arctan(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/((x^2+1)(x^2+4))

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

2 arctan (x) - arctan (\frac{x}{2}) + C

Solution steps

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

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

= 6 \cdot \int \frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 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

= 6 (\int \frac{1}{3 ( x ^{2} + 1 )} d x - \int \frac{1}{3 ( x ^{2} + 4 )} d x)

\int \frac{1}{3 ( x ^{2} + 1 )} d x = \frac{1}{3} arctan (x)

\int \frac{1}{3 ( x ^{2} + 4 )} d x = \frac{1}{6} arctan (\frac{x}{2})

= 6 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}))

Simplify

6 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2})) : 2 arctan (x) - arctan (\frac{x}{2})

= 2 arctan (x) - arctan (\frac{x}{2})

Add a constant to the solution

= 2 arctan (x) - arctan (\frac{x}{2}) + C

\int \frac{6 x ^{3}}{x ^{4} + 8 x ^{2} + 16} d x

\int \frac{8}{x ln ( x )} d x

\int \frac{2}{( 3 x + 5 ) ^{2}} d x

\int cot^{10} (x) csc^{4} (x) d x

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

What is the integral of 6/((x^2+1)(x^2+4)) ?
The integral of 6/((x^2+1)(x^2+4)) is 2arctan(x)-arctan(x/2)+C