What is the integral of (6x^2+5x+6)/((x^2+1)^2) ?

The integral of (6x^2+5x+6)/((x^2+1)^2) is 6arctan(x)-5/(2(x^2+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2+5x+6)/((x^2+1)^2)

\int \frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} d x

6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} : \frac{6}{x ^{2} + 1} + \frac{5 x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{6}{x ^{2} + 1} + \frac{5 x}{( x ^{2} + 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{6}{x ^{2} + 1} d x + \int \frac{5 x}{( x ^{2} + 1 ) ^{2}} d x

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

\int \frac{5 x}{( x ^{2} + 1 ) ^{2}} d x = - \frac{5}{2 ( x ^{2} + 1 )}

= 6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 6 arctan (x) - \frac{5}{2 ( x ^{2} + 1 )} + C

y^{^{'}} = \frac{3 ^{2} y ^{2} + 1}{3}, y (0) = - 0.5

\int_{0}^{5} π (25 - y^{2}) d y

\frac{\partial}{\partial x} (ln (x + 3 y + 5 z))

\int_{\frac{1}{2}}^{3} 12 x ln (2 x) d x

\int_{0}^{1} \frac{3}{x} d x

What is the integral of (6x^2+5x+6)/((x^2+1)^2) ?
The integral of (6x^2+5x+6)/((x^2+1)^2) is 6arctan(x)-5/(2(x^2+1))+C