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Popular >

integral of x/(x^2+16)

Solution

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

Solution

\frac{1}{2} ln 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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 16

= \frac{1}{2} ln x^{2} + 16

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 16 + C

Popular Examples

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\int \frac{x}{x - 21} d x

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\frac{d}{d x} ((x + 2)^{2} (x - 1))

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\int_{0}^{1} x^{2} + 1 d x

inverse oflaplace 1/(s((s-1)^2+1))inverse laplace

\frac{1}{s ( ( s - 1 ) ^{2} + 1 )}

Frequently Asked Questions (FAQ)

What is the integral of x/(x^2+16) ?
The integral of x/(x^2+16) is 1/2 ln|x^2+16|+C