What is the integral of sqrt(3x^2+4) ?

The integral of sqrt(3x^2+4) is 1/2 xsqrt(3x^2+4)+2/(sqrt(3))ln((|sqrt(3)x+sqrt(3x^2+4)|)/2)+C

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integral of sqrt(3x^2+4)

\int 3 x^{2} + 4 d x

\frac{1}{2} x 3 x^{2} + 4 + \frac{2}{3} ln (\frac{3 x + 3 x ^{2} + 4}{2}) + C

Solution steps

\int 3 x^{2} + 4 d x

Apply Trigonometric Substitution

: \int \frac{4}{3} sec^{3} (u) d u

= \int \frac{4}{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{4}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{4}{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{4}{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{4}{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{4}{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} + 4 + \frac{2}{3} ln (\frac{3 x + 3 x ^{2} + 4}{2})

= \frac{1}{2} x 3 x^{2} + 4 + \frac{2}{3} ln (\frac{3 x + 3 x ^{2} + 4}{2})

Add a constant to the solution

= \frac{1}{2} x 3 x^{2} + 4 + \frac{2}{3} ln (\frac{3 x + 3 x ^{2} + 4}{2}) + C

\int 1 ln (x^{2} - 1) d x

e^{x}, at x = 5

\frac{d}{d x} (x^{5} sin (x) tan (x))

x \to - 5 - lim (\frac{8}{( - 5 ) + 5})

\frac{d}{d x} ((3 x - 1)^{5})

What is the integral of sqrt(3x^2+4) ?
The integral of sqrt(3x^2+4) is 1/2 xsqrt(3x^2+4)+2/(sqrt(3))ln((|sqrt(3)x+sqrt(3x^2+4)|)/2)+C