What is the integral of 1/(2sqrt(x))arccos(sqrt(x)) ?

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

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integral of 1/(2sqrt(x))arccos(sqrt(x))

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

x arccos (x) - - x + 1 + C

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x} arccos (x) d x

Simplify

\frac{1}{x} arccos (x) : \frac{arccos ( x )}{x}

= \frac{1}{2} \cdot \int \frac{arccos ( x )}{x} d x

Apply Integration By Parts

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

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

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

= \frac{1}{2} (2 x arccos (x) - 2 1 - x)

Simplify

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

= x arccos (x) - - x + 1

Add a constant to the solution

= x arccos (x) - - x + 1 + C

\int \frac{3}{x - x} d x

\int x^{2} \cdot \frac{1}{4 x ^{2} + 9} d x

x^{2} + x + 1

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

\int \frac{1}{x - 1} d x

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