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

The integral of x/(x^2+64) is 1/2 ln|x^2+64|+C

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integral of x/(x^2+64)

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

\frac{1}{2} ln x^{2} + 64 + C

Solution steps

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 64

= \frac{1}{2} ln x^{2} + 64

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 64 + C

\frac{d y}{d x} = 7 y^{2} sin (x)

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

\frac{\partial}{\partial x} (x e^{2 x y})

x y^{2} \frac{d y}{d x} = y^{3} - x^{3}, y (1) = 1

f (x) - 2 x^{2}

What is the integral of x/(x^2+64) ?
The integral of x/(x^2+64) is 1/2 ln|x^2+64|+C