What is the integral of (arcsin(x))/((1-x^2)^{1/2)} ?

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

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integral of (arcsin(x))/((1-x^2)^{1/2)}

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

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

Solution steps

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

Apply Integration By Parts

: arcsin^{2} (x) - \int \frac{arcsin ( x )}{1 - x ^{2}} d x

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

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

= arcsin^{2} (x) - \frac{arcsin ^{2} ( x )}{2}

Add a constant to the solution

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

x \to 0 lim (x^{15 x})

\int \frac{x ^{3}}{6 - x ^{4}} d x

\frac{6 u ^{2}}{( u ^{2} + u ) ^{3}}

[δ (t + 1)]

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

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