What is the integral of 1/((196-x^2)^{3/2)} ?

The integral of 1/((196-x^2)^{3/2)} is x/(196(196-x^2)^{1/2)}+C

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integral of 1/((196-x^2)^{3/2)}

\int \frac{1}{( 196 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{196 ( 196 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 196 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{196 cos ^{2} ( u )} d u

= \int \frac{1}{196 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}{196} \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}{196} tan (u)

Substitute back

u = arcsin (\frac{1}{14} x)

= \frac{1}{196} tan (arcsin (\frac{1}{14} x))

Simplify

\frac{1}{196} tan (arcsin (\frac{1}{14} x)) : \frac{x}{196 ( 196 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{196 ( 196 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{196 ( 196 - x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{5 x - 12}{x ( x - 4 )} d x

\int \frac{( ln ( x ) ) ^{6}}{x} d x

t \cdot e^{- 4 t} cos (5 t)

cot^{2} (x)

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

What is the integral of 1/((196-x^2)^{3/2)} ?
The integral of 1/((196-x^2)^{3/2)} is x/(196(196-x^2)^{1/2)}+C