What is the integral of 10-x/((x^2+1)^2) ?

The integral of 10-x/((x^2+1)^2) is 10x+1/(2(x^2+1))+C

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integral of 10-x/((x^2+1)^2)

\int 10 - \frac{x}{( x ^{2} + 1 ) ^{2}} d x

10 x + \frac{1}{2 ( x ^{2} + 1 )} + C

Solution steps

\int 10 - \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 10 d x - \int \frac{x}{( x ^{2} + 1 ) ^{2}} d x

\int 10 d x = 10 x

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

= 10 x - (- \frac{1}{2 ( x ^{2} + 1 )})

Simplify

= 10 x + \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

= 10 x + \frac{1}{2 ( x ^{2} + 1 )} + C

What is the integral of 10-x/((x^2+1)^2) ?
The integral of 10-x/((x^2+1)^2) is 10x+1/(2(x^2+1))+C