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

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

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integral of 6/((1-x^2)^{3/2)}

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

\frac{6 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{6}{( 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

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

Apply Trigonometric Substitution

: \int sec^{2} (u) d u

= 6 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 6 tan (u)

Substitute back

u = arcsin (x)

= 6 tan (arcsin (x))

Simplify

6 tan (arcsin (x)) : \frac{6 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{6 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{6 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d y}{( 5 - 3 y ) ^{2}} = \frac{d x}{2 ( 4 - x ) ^{2}}

f^{^{'}} (x) = x e^{- x^{2}}

e^{- 5 t} \cdot e^{25 t} cos (20 t)

\int (1 - t^{2})^{3} d t

y = x^{- 3} + 5 x^{- 2}

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