Question

Evaluate $\int \frac{1}{(x^2+16)^{3/2}}dx$. Let x be the measure of the side opposite $\theta$. $\frac{x}{\sqrt{x^2+16}} + C$ $\frac{-1}{4\sqrt{x^2+16}} + C$ $\frac{-x}{16\sqrt{x^2+16}} + C$ $\frac{x}{16\sqrt{x^2+16}} + C$

          Evaluate $\int \frac{1}{(x^2+16)^{3/2}}dx$.
Let x be the measure of the side opposite $\theta$.
$\frac{x}{\sqrt{x^2+16}} + C$
$\frac{-1}{4\sqrt{x^2+16}} + C$
$\frac{-x}{16\sqrt{x^2+16}} + C$
$\frac{x}{16\sqrt{x^2+16}} + C$

Evaluate ∫(1)/((x^2+16)^3/2)dx.
Let x be the measure of the side opposite θ.
(x)/(√(x^2+16)) + C
(-1)/(4√(x^2+16)) + C
(-x)/(16√(x^2+16)) + C
(x)/(16√(x^2+16)) + C

Added by Laurie K.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ma. Theresa Alin

12/18/2021

Step 1

Let $u = x^2 + 16$, then $du = 2x dx$. Show more…

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Evaluate int (1)/((x^(2)+16)^((3)/(2)))dx Let x be the measure of the side opposite heta . (x)/(sqrt(x^(2)+16))+C (-1)/(4sqrt(x^(2)+16))+C (-x)/(16sqrt(x^(2)+16))+C (x)/(16sqrt(x^(2)+16))+C 1 Evaluate +163/2 Let x be the measure of the side opposite . o+ x+16 -1 +C 4x+16 16x+16 O 16x16