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

The integral of sqrt(16+9x^2) is 1/6 (3xsqrt(9x^2+16)+16ln((|3x+sqrt(9x^2+16)|)/4))+C

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integral of sqrt(16+9x^2)

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

\frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4})) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{16}{3} sec^{3} (u) d u

= \int \frac{16}{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{16}{3} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{16}{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{16}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

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

Simplify

\frac{16}{3} (\frac{sec ^{2} ( arctan ( \frac{3}{4} x ) ) sin ( arctan ( \frac{3}{4} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x))) : \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}))

= \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}))

Add a constant to the solution

= \frac{1}{6} (3 x 9 x^{2} + 16 + 16 ln (\frac{3 x + 9 x ^{2} + 16}{4})) + C

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}

f (x) = x + \frac{1}{x}

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

\int 5 cot^{3} (x) csc^{3} (x) d x

ln ∣ sec (x) ∣

What is the integral of sqrt(16+9x^2) ?
The integral of sqrt(16+9x^2) is 1/6 (3xsqrt(9x^2+16)+16ln((|3x+sqrt(9x^2+16)|)/4))+C