What is the integral of x/(x^2+81) ?

The integral of x/(x^2+81) is 1/2 ln|x^2+81|+C

Popular >

integral of x/(x^2+81)

\int \frac{x}{x ^{2} + 81} d x

\frac{1}{2} ln x^{2} + 81 + C

Solution steps

\int \frac{x}{x ^{2} + 81} 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} + 81

= \frac{1}{2} ln x^{2} + 81

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 81 + C

\frac{\partial}{\partial z} (5 zy + 2)

\int_{- 33}^{0} \frac{1}{16 - x} d x

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

\frac{\partial}{\partial x} (x^{2} y^{3} e^{x y})

x \to \infty lim (\frac{9 x - x ^{2} - 7 x ^{4}}{x ^{3} + 12 x})

What is the integral of x/(x^2+81) ?
The integral of x/(x^2+81) is 1/2 ln|x^2+81|+C