What is the integral of 1/((sqrt(u))(\sqrt{256-u))} ?

The integral of 1/((sqrt(u))(\sqrt{256-u))} is 2arcsin(1/16 sqrt(u))+C

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integral of 1/((sqrt(u))(\sqrt{256-u))}

\int \frac{1}{( u ) ( 256 - u )} d u

2 arcsin (\frac{1}{16} u) + C

Solution steps

\int \frac{1}{u 256 - u} d u

Apply u-substitution

: \int \frac{2}{256 - v ^{2}} d v

= \int \frac{2}{256 - v ^{2}} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{256 - v ^{2}} d v

Apply Trigonometric Substitution

: \int 1 d w

= 2 \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot w

Substitute back

= 2 \cdot 1 \cdot arcsin (\frac{1}{16} u)

Simplify

= 2 arcsin (\frac{1}{16} u)

Add a constant to the solution

= 2 arcsin (\frac{1}{16} u) + C

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What is the integral of 1/((sqrt(u))(\sqrt{256-u))} ?
The integral of 1/((sqrt(u))(\sqrt{256-u))} is 2arcsin(1/16 sqrt(u))+C