What is the integral of 8/((1-x^2)^{3/2)} ?

The integral of 8/((1-x^2)^{3/2)} is (8x)/((1-x^2)^{1/2)}+C

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integral of 8/((1-x^2)^{3/2)}

\int \frac{8}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{8 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{8}{( 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

= 8 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 8 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 8 tan (u)

Substitute back

u = arcsin (x)

= 8 tan (arcsin (x))

Simplify

8 tan (arcsin (x)) : \frac{8 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{8 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{8 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

f (x) = \frac{5}{x}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + x = 6 e^{- t}

(\frac{x + 12}{x - 12})^{5}

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

n = 0 \sum \infty \frac{0 ^{n}}{n !}

What is the integral of 8/((1-x^2)^{3/2)} ?
The integral of 8/((1-x^2)^{3/2)} is (8x)/((1-x^2)^{1/2)}+C