What is the integral of 1/(sqrt(x(4-x))) ?

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

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integral of 1/(sqrt(x(4-x)))

\int \frac{1}{x ( 4 - x )} d x

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

Solution steps

\int \frac{1}{x ( 4 - x )} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (4 - x) = x (4 - x)

= \int \frac{1}{x 4 - x} d x

Apply u-substitution

: \int \frac{2}{4 - u ^{2}} d u

= \int \frac{2}{4 - u ^{2}} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 2 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot v

Substitute back

= 2 \cdot 1 \cdot arcsin (\frac{1}{2} x)

Simplify

= 2 arcsin (\frac{1}{2} x)

Add a constant to the solution

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

sin^{3} (x)

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

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

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

\int \frac{ln ( 1 + e ^{x} )}{e ^{x}} d x

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