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

The integral of y/(sqrt(y^2+16)) is sqrt(y^2+16)+C

integral of y/(sqrt(y^2+16))

\int \frac{y}{y ^{2} + 16} d y

y^{2} + 16 + C

Solution steps

\int \frac{y}{y ^{2} + 16} d y

Apply u-substitution

= \int \frac{1}{2 u} d u

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = y^{2} + 16

= \frac{1}{2} \cdot 2 (y^{2} + 16)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (y^{2} + 16)^{\frac{1}{2}} : y^{2} + 16

= y^{2} + 16

Add a constant to the solution

= y^{2} + 16 + C

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What is the integral of y/(sqrt(y^2+16)) ?
The integral of y/(sqrt(y^2+16)) is sqrt(y^2+16)+C