What is the integral of x/(x^2+36) ?

The integral of x/(x^2+36) is 1/2 ln|x^2+36|+C

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integral of x/(x^2+36)

\int \frac{x}{x ^{2} + 36} d x

\frac{1}{2} ln x^{2} + 36 + C

Solution steps

\int \frac{x}{x ^{2} + 36} 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} + 36

= \frac{1}{2} ln x^{2} + 36

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 36 + C

\frac{\partial}{\partial x} ((3 + x^{2} - 3 y^{2})^{\frac{1}{2}})

2 x (\frac{d y}{d x}) - 10 y = x^{- 6}

x^{2} e^{3 x}

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = b, y^{^{'}} (0) = 8

cot^{2} (sin (x))

What is the integral of x/(x^2+36) ?
The integral of x/(x^2+36) is 1/2 ln|x^2+36|+C