What is the integral of+x/(3x^2+1) ?

The integral of+x/(3x^2+1) is 1/6 ln|3x^2+1|+C

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integral of+x/(3x^2+1)

\int + \frac{x}{3 x ^{2} + 1} d x

\frac{1}{6} ln 3 x^{2} + 1 + C

Solution steps

\int \frac{x}{3 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} + 1

= \frac{1}{6} ln 3 x^{2} + 1

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} + 1 + C

\frac{d}{d x} ((2 x^{3} + 5)^{10})

\int sec (x) d x

\frac{d}{d x} (\frac{e ^{x}}{x} + ln (x))

\int \frac{u}{1 - u ^{2}} d u

\frac{d}{d x} (\frac{csc ^{3} ( x )}{ln ( x ^{2} )})

What is the integral of+x/(3x^2+1) ?
The integral of+x/(3x^2+1) is 1/6 ln|3x^2+1|+C