What is the integral of sqrt(4+9x^2) ?

The integral of sqrt(4+9x^2) is 1/6 (3xsqrt(9x^2+4)+4ln((|3x+sqrt(9x^2+4)|)/2))+C

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integral of sqrt(4+9x^2)

\int 4 + 9 x^{2} d x

\frac{1}{6} (3 x 9 x^{2} + 4 + 4 ln (\frac{3 x + 9 x ^{2} + 4}{2})) + C

Solution steps

\int 4 + 9 x^{2} 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}{6} (3 x 9 x^{2} + 4 + 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}))

= \frac{1}{6} (3 x 9 x^{2} + 4 + 4 ln (\frac{3 x + 9 x ^{2} + 4}{2}))

Add a constant to the solution

= \frac{1}{6} (3 x 9 x^{2} + 4 + 4 ln (\frac{3 x + 9 x ^{2} + 4}{2})) + C

\int sin (3 x) cos (4 x) sin (2 x) d x

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{x - π})

\frac{r}{r ^{2} + 7}

f (x) = \frac{4 x ^{3} + 2 x + 5}{x}

f (x) = \frac{5}{( x + 2 ) ^{2}}

What is the integral of sqrt(4+9x^2) ?
The integral of sqrt(4+9x^2) is 1/6 (3xsqrt(9x^2+4)+4ln((|3x+sqrt(9x^2+4)|)/2))+C