What is the integral of sqrt(9x^2+81) ?

The integral of sqrt(9x^2+81) is 27(1/18 xsqrt(x^2+9)+1/2 ln((|x+sqrt(9+x^2)|)/3))+C

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integral of sqrt(9x^2+81)

\int 9 x^{2} + 81 d x

27 (\frac{1}{18} x x^{2} + 9 + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3})) + C

Solution steps

\int 9 x^{2} + 81 d x

Apply Trigonometric Substitution

: \int 27 sec^{3} (u) d u

= \int 27 sec^{3} (u) d u

Take the constant out:

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

= 27 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 27 (\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) ∣

= 27 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{3}{9} x)

= 27 (\frac{sec ^{2} ( arctan ( \frac{3}{9} x ) ) sin ( arctan ( \frac{3}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{9} x)) + sec (arctan (\frac{3}{9} x)))

Simplify

27 (\frac{sec ^{2} ( arctan ( \frac{3}{9} x ) ) sin ( arctan ( \frac{3}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{3}{9} x)) + sec (arctan (\frac{3}{9} x))) : 27 (\frac{1}{18} x x^{2} + 9 + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3}))

= 27 (\frac{1}{18} x x^{2} + 9 + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3}))

Add a constant to the solution

= 27 (\frac{1}{18} x x^{2} + 9 + \frac{1}{2} ln (\frac{x + 9 + x ^{2}}{3})) + C

\int 6 x (3 x^{2} - 1) d x

\frac{1}{x ^{2} + 9}

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\int cos (2 x) cos (24 x) d x

\int \frac{e ^{8 x}}{e ^{8 x} - 9 e ^{- 8 x}} d x

What is the integral of sqrt(9x^2+81) ?
The integral of sqrt(9x^2+81) is 27(1/18 xsqrt(x^2+9)+1/2 ln((|x+sqrt(9+x^2)|)/3))+C