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

The integral of (x+sqrt(1-x^2))^2 is -2/3 (-x^2+1)^{3/2}+x+C

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integral of (x+sqrt(1-x^2))^2

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

- \frac{2}{3} (- x^{2} + 1)^{\frac{3}{2}} + x + C

Solution steps

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

Expand

(x + 1 - x^{2})^{2} : 2 x - x^{2} + 1 + 1

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

\int 2 x - x^{2} + 1 d x = - \frac{2}{3} (- x^{2} + 1)^{\frac{3}{2}}

\int 1 d x = x

= - \frac{2}{3} (- x^{2} + 1)^{\frac{3}{2}} + x

Add a constant to the solution

= - \frac{2}{3} (- x^{2} + 1)^{\frac{3}{2}} + x + C

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What is the integral of (x+sqrt(1-x^2))^2 ?
The integral of (x+sqrt(1-x^2))^2 is -2/3 (-x^2+1)^{3/2}+x+C