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

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

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integral of sqrt(9t^2+4)

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

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

Solution steps

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

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} t)

= \frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} t ) ) sin ( arctan ( \frac{3}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} t)) + sec (arctan (\frac{3}{2} t)))

Simplify

\frac{4}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{2} t ) ) sin ( arctan ( \frac{3}{2} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{2} t)) + sec (arctan (\frac{3}{2} t))) : \frac{1}{6} (3 t 9 t^{2} + 4 + 4 ln (\frac{3 t + 9 t ^{2} + 4}{2}))

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

Add a constant to the solution

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

y^{^{'}} = x + y + 1

\frac{d}{d x} (\frac{1 + 2 x}{3 - 4 x})

\int cos^{n} (x) d x

25 y^{^{''}} + 10 y^{^{'}} + y = 0

\frac{\partial}{\partial x} (\frac{9}{x ^{2} + y ^{2}})

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