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

\frac{\partial}{\partial y} (x^{2} ln (x y))

f (x) = 4 x - 3 x, (4, 10)

\frac{d}{d x} (e^{x + l n (x)})

x \to \infty lim ((- 2)^{x})

\frac{\partial}{\partial x} (2 π x y + π x^{2})

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