What is the integral of x/8 sqrt(64+x^2) ?

The integral of x/8 sqrt(64+x^2) is 1/24 (64+x^2)^{3/2}+C

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integral of x/8 sqrt(64+x^2)

\int \frac{x}{8} 64 + x^{2} d x

\frac{1}{24} (64 + x^{2})^{\frac{3}{2}} + C

Solution steps

\int \frac{x}{8} 64 + x^{2} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int x 64 + x^{2} d x

Apply u-substitution

: \int u^{2} d u

= \frac{1}{8} \cdot \int u^{2} d u

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{u ^{3}}{3}

Substitute back

u = 64 + x^{2}

= \frac{1}{8} \cdot \frac{( 64 + x ^{2} ) ^{3}}{3}

Simplify

\frac{1}{8} \cdot \frac{( 64 + x ^{2} ) ^{3}}{3} : \frac{1}{24} (64 + x^{2})^{\frac{3}{2}}

= \frac{1}{24} (64 + x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{24} (64 + x^{2})^{\frac{3}{2}} + C

y = x^{2} + 2, (1, 3)

x \to 0 lim (\frac{1}{x ^{2} ( x - 7 )})

\frac{d}{d x} (ln (y) + y^{2} + cos (x))

y^{^{''}} + 2 y^{^{'}} + y = sin (2 t)

What is the integral of x/8 sqrt(64+x^2) ?
The integral of x/8 sqrt(64+x^2) is 1/24 (64+x^2)^{3/2}+C