What is the integral of x^3arcsin(1/x) ?

The integral of x^3arcsin(1/x) is 1/4 x^4arcsin(1/x)+1/4 (1/3 (x^2-1)^{3/2}+sqrt(x^2-1))+C

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integral of x^3arcsin(1/x)

\int x^{3} arcsin (\frac{1}{x}) d x

\frac{1}{4} x^{4} arcsin (\frac{1}{x}) + \frac{1}{4} (\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1) + C

Solution steps

\int x^{3} arcsin (\frac{1}{x}) d x

Apply Integration By Parts

: \frac{1}{4} x^{4} arcsin (\frac{1}{x}) - \int - \frac{x ^{2} x ^{2}}{4 x ^{2} - 1} d x

= \frac{1}{4} x^{4} arcsin (\frac{1}{x}) - \int - \frac{x ^{2} x ^{2}}{4 x ^{2} - 1} d x

\int - \frac{x ^{2} x ^{2}}{4 x ^{2} - 1} d x = - \frac{1}{4} (\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1)

= \frac{1}{4} x^{4} arcsin (\frac{1}{x}) - (- \frac{1}{4} (\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1))

Simplify

= \frac{1}{4} x^{4} arcsin (\frac{1}{x}) + \frac{1}{4} (\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1)

Add a constant to the solution

= \frac{1}{4} x^{4} arcsin (\frac{1}{x}) + \frac{1}{4} (\frac{1}{3} (x^{2} - 1)^{\frac{3}{2}} + x^{2} - 1) + C

\frac{\partial}{\partial x} (y^{3} - x^{2} y)

f (x) = 8 x^{9} - 3 x^{6} + 12 x^{3}

y = 4 x - x^{2}, y = 0

(X - 1) e^{X}

\frac{d y}{d x}, 4 y^{2} + 2 = 3 x^{2}

What is the integral of x^3arcsin(1/x) ?
The integral of x^3arcsin(1/x) is 1/4 x^4arcsin(1/x)+1/4 (1/3 (x^2-1)^{3/2}+sqrt(x^2-1))+C