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

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

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integral of x/(3+x^2)

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

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

Solution steps

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

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

Add a constant to the solution

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

x y^{^{'}} - 2 y = x^{3}

y = x - 1, y = 0, x = 9

h \to 1 lim (\frac{e ^{- h} - 1}{h})

a^{2} + x^{2}

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