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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

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

x \to 0 lim ((sin (x))^{x})

\frac{d}{d x} (x^{c o s (2 π x)})

\frac{d}{d t} (\frac{4 t ^{2} + 7 t ^{5}}{t})

\int x \cdot sin (nπ x) d x

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