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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2 sec ( u )} d u

= - 2 \cdot \int \frac{1}{2 sec ( u )} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{2} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= - 2 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= - 2 \cdot \frac{1}{2} sin (u)

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int - 9 tan^{2} (x) d x

f (x) = x, at x = 16

\frac{d}{d x} (\frac{1}{1 + x ^{12}})

\frac{d}{d x} (1 - e^{- 8 x})

2 x + 5 x^{- 1}

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