What is the integral of ln(x)arcsin(x) ?

The integral of ln(x)arcsin(x) is

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integral of ln(x)arcsin(x)

\int ln (x) arcsin (x) d x

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

Solution steps

\int ln (x) arcsin (x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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