What is the integral of x/(x^2+3) ?

The integral of x/(x^2+3) is 1/2 ln|x^2+3|+C

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integral of x/(x^2+3)

\int \frac{x}{x ^{2} + 3} d x

\frac{1}{2} ln x^{2} + 3 + C

Solution steps

\int \frac{x}{x ^{2} + 3} 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} + 3

= \frac{1}{2} ln x^{2} + 3

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 3 + C

\int \frac{x ^{3}}{x ^{2} - 49} d x

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

x \to 2 lim (x^{4} + 5^{x} - 2)

x \to - 1 lim (2 x^{3} + 3 x^{2} - 6)

\int_{- 1}^{3} x^{2} - 2 x d x

What is the integral of x/(x^2+3) ?
The integral of x/(x^2+3) is 1/2 ln|x^2+3|+C