What is the integral of sqrt(8+3x^2) ?

The integral of sqrt(8+3x^2) is 1/2 xsqrt(3x^2+8)+2/(sqrt(3))(2ln|sqrt(3)x+sqrt(3x^2+8)|-3ln(2))+C

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integral of sqrt(8+3x^2)

\int 8 + 3 x^{2} d x

\frac{1}{2} x 3 x^{2} + 8 + \frac{2}{3} (2 ln 3 x + 3 x^{2} + 8 - 3 ln (2)) + C

Solution steps

\int 8 + 3 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{8}{3} sec^{3} (u) d u

= \int \frac{8}{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{8}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{8}{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{8}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= \frac{8}{3} \frac{sec ^{2} ( arctan ( \frac{3}{2 2} x ) ) sin ( arctan ( \frac{3}{2 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2 2} x)) + sec (arctan (\frac{3}{2 2} x))

Simplify

\frac{8}{3} \frac{sec ^{2} ( arctan ( \frac{3}{2 2} x ) ) sin ( arctan ( \frac{3}{2 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2 2} x)) + sec (arctan (\frac{3}{2 2} x)) : \frac{1}{2} x 3 x^{2} + 8 + \frac{2}{3} (2 ln 3 x + 3 x^{2} + 8 - 3 ln (2))

= \frac{1}{2} x 3 x^{2} + 8 + \frac{2}{3} (2 ln 3 x + 3 x^{2} + 8 - 3 ln (2))

Add a constant to the solution

= \frac{1}{2} x 3 x^{2} + 8 + \frac{2}{3} (2 ln 3 x + 3 x^{2} + 8 - 3 ln (2)) + C

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What is the integral of sqrt(8+3x^2) ?
The integral of sqrt(8+3x^2) is 1/2 xsqrt(3x^2+8)+2/(sqrt(3))(2ln|sqrt(3)x+sqrt(3x^2+8)|-3ln(2))+C