What is the integral of x/4 sqrt(16+x^2) ?

The integral of x/4 sqrt(16+x^2) is 1/12 (16+x^2)^{3/2}+C

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integral of x/4 sqrt(16+x^2)

\int \frac{x}{4} 16 + x^{2} d x

\frac{1}{12} (16 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int \frac{x}{4} 16 + x^{2} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int x 16 + x^{2} d x

Apply u-substitution

: \int u^{2} d u

= \frac{1}{4} \cdot \int u^{2} d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{u ^{3}}{3}

Substitute back

u = 16 + x^{2}

= \frac{1}{4} \cdot \frac{( 16 + x ^{2} ) ^{3}}{3}

Simplify

\frac{1}{4} \cdot \frac{( 16 + x ^{2} ) ^{3}}{3} : \frac{1}{12} (16 + x^{2})^{\frac{3}{2}}

= \frac{1}{12} (16 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (16 + x^{2})^{\frac{3}{2}} + C

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What is the integral of x/4 sqrt(16+x^2) ?
The integral of x/4 sqrt(16+x^2) is 1/12 (16+x^2)^{3/2}+C