What is the integral of (8+x^2)^{-3/2} ?

The integral of (8+x^2)^{-3/2} is x/(8(8+x^2)^{1/2)}+C

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integral of (8+x^2)^{-3/2}

\int (8 + x^{2})^{- \frac{3}{2}} d x

\frac{x}{8 ( 8 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int (8 + x^{2})^{- \frac{3}{2}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 8 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{1}{8 sec ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{8} sin (u)

Substitute back

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

= \frac{1}{8} sin (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x))

Simplify

\frac{1}{8} sin (arctan (\frac{1}{2 \cdot 2 ^{\frac{1}{2}}} x)) : \frac{x}{8 ( 8 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{8 ( 8 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{8 ( 8 + x ^{2} ) ^{\frac{1}{2}}} + C

What is the integral of (8+x^2)^{-3/2} ?
The integral of (8+x^2)^{-3/2} is x/(8(8+x^2)^{1/2)}+C