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

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{5 + x}{5 - x} d x

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

t an g e n t f (x) = \frac{4}{x + 1}, at x = 1

x \to 25 lim (x - 25)

(x + 1) d y = (x + 6) d x

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