What is the integral of 1/(sqrt(x)\sqrt{9-x)} ?

The integral of 1/(sqrt(x)\sqrt{9-x)} is 2arcsin(1/3 sqrt(x))+C

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integral of 1/(sqrt(x)\sqrt{9-x)}

\int \frac{1}{x 9 - x} d x

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

Solution steps

\int \frac{1}{x 9 - x} d x

Apply u-substitution

: \int \frac{2}{9 - u ^{2}} d u

= \int \frac{2}{9 - 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}{9 - 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}{3} x)

Simplify

= 2 arcsin (\frac{1}{3} x)

Add a constant to the solution

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

\frac{d}{d x} (\frac{2}{( 3 - x ) ^{3}})

\int \frac{x ^{4} - 2 x ^{2} + 7}{x ^{4}} d x

\int_{0}^{1} \frac{x ^{4} + 8}{x} d x

\int x e^{x^{2} + 1} d x

(1 + 8 x^{2}) (x - x^{2})

What is the integral of 1/(sqrt(x)\sqrt{9-x)} ?
The integral of 1/(sqrt(x)\sqrt{9-x)} is 2arcsin(1/3 sqrt(x))+C