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

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

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integral of sqrt(3x^2+2)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{2}{3} sec^{3} (u) d u

= \int \frac{2}{3} sec^{3} (u) d u

Take the constant out:

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

= \frac{2}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{2}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{2}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{3}{2} x)

= \frac{2}{3} \frac{sec ^{2} ( arctan ( \frac{3}{2} x ) ) sin ( arctan ( \frac{3}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x))

Simplify

\frac{2}{3} \frac{sec ^{2} ( arctan ( \frac{3}{2} x ) ) sin ( arctan ( \frac{3}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} x)) + sec (arctan (\frac{3}{2} x)) : \frac{1}{2} x 3 x^{2} + 2 + \frac{1}{3} ln \frac{3}{2} x + \frac{2 + 3 x ^{2}}{2}

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

Add a constant to the solution

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

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