What is the integral of (-5)/((x^2+5)^{3/2)} ?

The integral of (-5)/((x^2+5)^{3/2)} is -x/((5+x^2)^{1/2)}+C

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integral of (-5)/((x^2+5)^{3/2)}

\int \frac{- 5}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

- \frac{x}{( 5 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{- 5}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= - 5 \cdot \int \frac{1}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= - 5 \cdot \int \frac{1}{5 sec ( u )} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{5} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= - 5 \cdot \frac{1}{5} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - 5 \cdot \frac{1}{5} sin (u)

Substitute back

u = arctan (\frac{1}{5 ^{\frac{1}{2}}} x)

= - 5 \cdot \frac{1}{5} sin (arctan (\frac{1}{5 ^{\frac{1}{2}}} x))

Simplify

- 5 \cdot \frac{1}{5} sin (arctan (\frac{1}{5 ^{\frac{1}{2}}} x)) : - \frac{x}{( 5 + x ^{2} ) ^{\frac{1}{2}}}

= - \frac{x}{( 5 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x}{( 5 + x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d}{d t} (\frac{6}{5} - t^{2})

a re a \frac{x ^{2}}{8}, 2 x^{2}, 2, 0

n = 1 \sum \infty n! x^{n}

\frac{d}{d y} (x y 1 - \frac{x ^{2}}{4} - \frac{y ^{2}}{9})

x \to 0 lim (\frac{x ^{2} - 3 x}{2 x})

What is the integral of (-5)/((x^2+5)^{3/2)} ?
The integral of (-5)/((x^2+5)^{3/2)} is -x/((5+x^2)^{1/2)}+C