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

The integral of (2x^2)/(1+sqrt(1-x^2)) is 2x-arcsin(x)-1/2 sin(2arcsin(x))+C

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

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

2 x - arcsin (x) - \frac{1}{2} sin (2 arcsin (x)) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Multiply by the conjugate of

1 + 1 - x^{2} : \frac{x ^{2}}{1 - 1 - x ^{2}}

= 2 \cdot \int \frac{x ^{2}}{\frac{x ^{2}}{1 - 1 - x ^{2}}} d x

Simplify

\frac{x ^{2}}{\frac{x ^{2}}{1 - 1 - x ^{2}}} : 1 - - x^{2} + 1

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

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

\int 1 d x = x

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

= 2 (x - \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x))))

Simplify

2 (x - \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))) : 2 x - arcsin (x) - \frac{1}{2} sin (2 arcsin (x))

= 2 x - arcsin (x) - \frac{1}{2} sin (2 arcsin (x))

Add a constant to the solution

= 2 x - arcsin (x) - \frac{1}{2} sin (2 arcsin (x)) + C

\int \frac{1}{a ^{x}} d x

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

\int sec (\frac{x}{9}) d x

\int cos (x) ln (tan (x)) d x

\int \frac{1}{1 + x ^{2}} x d x

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