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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{a} tan (u)

Substitute back

u = arcsin (a x)

= \frac{1}{a} tan (arcsin (a x))

Simplify

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

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

Add a constant to the solution

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

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\int \frac{7}{x ^{4} - 1} d x

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

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

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