What is the integral of sqrt(9+16t^2) ?

The integral of sqrt(9+16t^2) is 1/8 (4tsqrt(16t^2+9)+9ln((|4t+sqrt(16t^2+9)|)/3))+C

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integral of sqrt(9+16t^2)

\int 9 + 16 t^{2} d t

\frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3})) + C

Solution steps

\int 9 + 16 t^{2} d t

Apply Trigonometric Substitution

: \int \frac{9}{4} sec^{3} (u) d u

= \int \frac{9}{4} sec^{3} (u) d u

Take the constant out:

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

= \frac{9}{4} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{9}{4} (\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{9}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{4}{3} t)

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

Simplify

\frac{9}{4} (\frac{sec ^{2} ( arctan ( \frac{4}{3} t ) ) sin ( arctan ( \frac{4}{3} t ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{4}{3} t)) + sec (arctan (\frac{4}{3} t))) : \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3}))

= \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3}))

Add a constant to the solution

= \frac{1}{8} (4 t 16 t^{2} + 9 + 9 ln (\frac{4 t + 16 t ^{2} + 9}{3})) + C

\int 1 - cos (8 x) d x

x \to 3 lim (\frac{1 - x}{x - 3})

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

\frac{d}{d x} (x^{n} (ln (x))^{2})

f (x) = \frac{9 - x e ^{x}}{x + e ^{x}}

What is the integral of sqrt(9+16t^2) ?
The integral of sqrt(9+16t^2) is 1/8 (4tsqrt(16t^2+9)+9ln((|4t+sqrt(16t^2+9)|)/3))+C