What is the integral of xsqrt(16+x^2) ?

The integral of xsqrt(16+x^2) is 1/3 (16+x^2)^{3/2}+C

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integral of xsqrt(16+x^2)

\int x 16 + x^{2} d x

\frac{1}{3} (16 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 16 + x^{2} d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = 16 + x^{2}

= \frac{( 16 + x ^{2} ) ^{3}}{3}

Simplify

\frac{( 16 + x ^{2} ) ^{3}}{3} : \frac{1}{3} (16 + x^{2})^{\frac{3}{2}}

= \frac{1}{3} (16 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (16 + x^{2})^{\frac{3}{2}} + C

f (x) = ln (1 + \frac{1}{x})

\frac{d}{d x} (3 x sin (π x))

(\frac{x}{x ^{2} + 1})^{^{'}}

\frac{x}{x ^{2} + 8}

\int_{- \infty}^{2} x e^{x} d x

What is the integral of xsqrt(16+x^2) ?
The integral of xsqrt(16+x^2) is 1/3 (16+x^2)^{3/2}+C