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

The integral of (xarccos(x))/(sqrt(1-x^2)) is -arccos(x)sqrt(1-x^2)-x+C

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

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

- arccos (x) 1 - x^{2} - x + C

Solution steps

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

Apply u-substitution

: \int - cos (u) u d u

= \int - cos (u) u d u

Take the constant out:

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

= - \int cos (u) u d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

= - (u sin (u) - \int sin (u) d u)

\int sin (u) d u = - cos (u)

= - (u sin (u) - (- cos (u)))

Substitute back

u = arccos (x)

= - (arccos (x) sin (arccos (x)) - (- cos (arccos (x))))

Simplify

- (arccos (x) sin (arccos (x)) - (- cos (arccos (x)))) : - arccos (x) 1 - x^{2} - x

= - arccos (x) 1 - x^{2} - x

Add a constant to the solution

= - arccos (x) 1 - x^{2} - x + C

x^{2} y^{^{''}} + 6 x y^{^{'}} + 6 y = 0

\int_{1}^{\infty} e^{- x} d x

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

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{3}{x ^{3}})

\frac{d}{d x} (∣ x ∣ + ∣ x - 1 ∣)

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