What is the integral of (16)/((16-x^2)^{3/2)} ?

The integral of (16)/((16-x^2)^{3/2)} is x/((16-x^2)^{1/2)}+C

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integral of (16)/((16-x^2)^{3/2)}

\int \frac{16}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{16}{( 16 - 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

= 16 \cdot \int \frac{1}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 cos ^{2} ( u )} d u

= 16 \cdot \int \frac{1}{16 cos ^{2} ( u )} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

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

= 16 \cdot \frac{1}{16} tan (u)

Substitute back

u = arcsin (\frac{1}{4} x)

= 16 \cdot \frac{1}{16} tan (arcsin (\frac{1}{4} x))

Simplify

16 \cdot \frac{1}{16} tan (arcsin (\frac{1}{4} x)) : \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

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What is the integral of (16)/((16-x^2)^{3/2)} ?
The integral of (16)/((16-x^2)^{3/2)} is x/((16-x^2)^{1/2)}+C