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

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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

\frac{d S}{d r} = k S

\int sin^{3} (u) d u

\frac{d}{d x} ((ln (x))^{- 1})

y = x^{\frac{4}{3}}

\frac{1}{s ^{2} - 1}

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