What is the integral of 4 1/((4-x^2)^{3/2)} ?

The integral of 4 1/((4-x^2)^{3/2)} is x/((4-x^2)^{1/2)}+C

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integral of 4 1/((4-x^2)^{3/2)}

\int 4 \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int 4 \cdot \frac{1}{( 4 - 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

= 4 \cdot \int \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4 cos ^{2} ( u )} d u

= 4 \cdot \int \frac{1}{4 cos ^{2} ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

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

= 4 \cdot \frac{1}{4} tan (u)

Substitute back

u = arcsin (\frac{1}{2} x)

= 4 \cdot \frac{1}{4} tan (arcsin (\frac{1}{2} x))

Simplify

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

= \frac{x}{( 4 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{7 sin ( x )}{cos ^{3} ( x )} d x

\int \frac{1}{x ln ( x ^{9} )} d x

\int_{0}^{1} \frac{2}{2 - x ^{2}} d x

y = 3 x - 1

f (x) = e^{x} (x + x x)

What is the integral of 4 1/((4-x^2)^{3/2)} ?
The integral of 4 1/((4-x^2)^{3/2)} is x/((4-x^2)^{1/2)}+C