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

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

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integral of sqrt(9+16x^2)

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

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

Solution steps

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

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} x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{3 x - 2 y}{4 x ^{2} + y})

\int_{0}^{l n (5)} e^{x} d x

\frac{d}{d x} (34 (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x))

\int sin (ln (4 x)) d x

\frac{d}{d x} (e^{3 x} cos (2 x))

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