What is the integral of 6/((9-x^2)^{3/2)} ?

The integral of 6/((9-x^2)^{3/2)} is (2x)/(3(-x^2+9)^{1/2)}+C

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integral of 6/((9-x^2)^{3/2)}

\int \frac{6}{( 9 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{2 x}{3 ( - x ^{2} + 9 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{6}{( 9 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 9 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{9 cos ^{2} ( u )} d u

= 6 \cdot \int \frac{1}{9 cos ^{2} ( u )} d u

Take the constant out:

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

= 6 \cdot \frac{1}{9} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= 6 \cdot \frac{1}{9} tan (u)

Substitute back

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

= 6 \cdot \frac{1}{9} tan (arcsin (\frac{1}{3} x))

Simplify

6 \cdot \frac{1}{9} tan (arcsin (\frac{1}{3} x)) : \frac{2 x}{3 ( - x ^{2} + 9 ) ^{\frac{1}{2}}}

= \frac{2 x}{3 ( - x ^{2} + 9 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{2 x}{3 ( - x ^{2} + 9 ) ^{\frac{1}{2}}} + C

\int \frac{3 x}{9 x ^{2} + 1} d x

\int \frac{5}{x - 2} d x

\int (x + 19) 38 x + x^{2} d x

\int \frac{7}{5 x ^{2} + 11} d x

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

What is the integral of 6/((9-x^2)^{3/2)} ?
The integral of 6/((9-x^2)^{3/2)} is (2x)/(3(-x^2+9)^{1/2)}+C