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

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

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

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

x^{2} + 16 + C

Solution steps

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

Apply u-substitution

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

= \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

: 2 u^{\frac{1}{2}}

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

Substitute back

u = x^{2} + 16

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

Simplify

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

= x^{2} + 16

Add a constant to the solution

= x^{2} + 16 + C

\frac{\partial}{\partial y} (e^{5 x y})

\int 9^{x} d x

f (x) = 0.14 x^{2} + 0.68 x + 3.1

y = sin (4 x), (0, \frac{π}{4})

y^{^{'}} = 2 x + y, y (0) = 5

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