What is the integral of ((7x^2+x+7))/((x^2+1)^2) ?

The integral of ((7x^2+x+7))/((x^2+1)^2) is 7arctan(x)-1/(2(x^2+1))+C

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integral of ((7x^2+x+7))/((x^2+1)^2)

\int \frac{( 7 x ^{2} + x + 7 )}{( x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{7 x ^{2} + x + 7}{( x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{7 x ^{2} + x + 7}{( x ^{2} + 1 ) ^{2}} : \frac{7}{x ^{2} + 1} + \frac{x}{( x ^{2} + 1 ) ^{2}}

= \int \frac{7}{x ^{2} + 1} + \frac{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{7}{x ^{2} + 1} d x + \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x

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

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

= 7 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 7 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )} + C

\int \frac{x ^{3} + 1}{x ^{2} - 4} d x

\int_{0}^{4} e^{- x} d x

n = 1 \sum \infty (\frac{1}{4})^{n}

f (x) = x^{4} + 5 x

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

What is the integral of ((7x^2+x+7))/((x^2+1)^2) ?
The integral of ((7x^2+x+7))/((x^2+1)^2) is 7arctan(x)-1/(2(x^2+1))+C