What is the integral of 1/((25-x^2)^{3/2)} ?

The integral of 1/((25-x^2)^{3/2)} is x/(25(25-x^2)^{1/2)}+C

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integral of 1/((25-x^2)^{3/2)}

\int \frac{1}{( 25 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{25 ( 25 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 25 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{25 cos ^{2} ( u )} d u

= \int \frac{1}{25 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}{25} \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}{25} tan (u)

Substitute back

u = arcsin (\frac{1}{5} x)

= \frac{1}{25} tan (arcsin (\frac{1}{5} x))

Simplify

\frac{1}{25} tan (arcsin (\frac{1}{5} x)) : \frac{x}{25 ( 25 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{25 ( 25 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{25 ( 25 - x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{2 x - 3}{x ^{2} 6 x ^{13}} d x

\int \frac{e ^{2 x}}{e ^{x} + 1} d x

x \to 0 lim (\frac{( x + 2 ) ^{2} - 8}{x})

x^{2} + 8 x - 3, - x^{2} + 4 x + 3

n = 1 \sum \infty \frac{5 ^{n}}{9}

What is the integral of 1/((25-x^2)^{3/2)} ?
The integral of 1/((25-x^2)^{3/2)} is x/(25(25-x^2)^{1/2)}+C