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

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

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integral of xsqrt(x^2+16)

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int - x cos (x^{2}) d x

\frac{d}{d θ} (cos (2 θ))

(x^{2} + 1) (\frac{d y}{d x}) + x y - x = 0

\frac{d}{d x} (\frac{1}{3} \cdot x^{3})

f (x) = e^{\frac{x}{( x - 9 )}}

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