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

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

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integral of 9/((1-x^2)^{3/2)}

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

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

Solution steps

\int \frac{9}{( 1 - 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

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

Apply Trigonometric Substitution

= 9 \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 9 tan (u)

Substitute back

u = arcsin (x)

= 9 tan (arcsin (x))

Simplify

9 tan (arcsin (x)) : \frac{9 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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