What is the integral of 1/((144-x^2)^{3/2)} ?

The integral of 1/((144-x^2)^{3/2)} is x/(144(144-x^2)^{1/2)}+C

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integral of 1/((144-x^2)^{3/2)}

\int \frac{1}{( 144 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{144 ( 144 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 144 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{144 cos ^{2} ( u )} d u

= \int \frac{1}{144 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}{144} \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}{144} tan (u)

Substitute back

u = arcsin (\frac{1}{12} x)

= \frac{1}{144} tan (arcsin (\frac{1}{12} x))

Simplify

\frac{1}{144} tan (arcsin (\frac{1}{12} x)) : \frac{x}{144 ( 144 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{144 ( 144 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{144 ( 144 - x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (e^{- x - y})

\int y^{2} cos (x) + cos (y) d x

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

\frac{\partial}{\partial x} (x cos (θ))

f (x) = 5 e^{x} x

What is the integral of 1/((144-x^2)^{3/2)} ?
The integral of 1/((144-x^2)^{3/2)} is x/(144(144-x^2)^{1/2)}+C