What is the integral of (x^2)/(sqrt(16x^2+81)) ?

The integral of (x^2)/(sqrt(16x^2+81)) is 1/128 (-81ln((|4x+sqrt(16x^2+81)|)/9)+4xsqrt(16x^2+81))+C

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integral of (x^2)/(sqrt(16x^2+81))

\int \frac{x ^{2}}{16 x ^{2} + 81} d x

\frac{1}{128} (- 81 ln (\frac{4 x + 16 x ^{2} + 81}{9}) + 4 x 16 x^{2} + 81) + C

Solution steps

\int \frac{x ^{2}}{16 x ^{2} + 81} d x

Apply Trigonometric Substitution

: \int \frac{81 tan ^{2} ( u ) sec ( u )}{64} d u

= \int \frac{81 tan ^{2} ( u ) sec ( u )}{64} d u

Take the constant out:

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

= \frac{81}{64} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{81}{64} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \frac{81}{64} \cdot \int - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{81}{64} (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

= \frac{81}{64} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{4}{9} x)

= \frac{81}{64} (- ln tan (arctan (\frac{4}{9} x)) + sec (arctan (\frac{4}{9} x)) + \frac{1}{2} sec (arctan (\frac{4}{9} x)) tan (arctan (\frac{4}{9} x)) + \frac{1}{2} ln tan (arctan (\frac{4}{9} x)) + sec (arctan (\frac{4}{9} x)))

Simplify

\frac{81}{64} (- ln tan (arctan (\frac{4}{9} x)) + sec (arctan (\frac{4}{9} x)) + \frac{1}{2} sec (arctan (\frac{4}{9} x)) tan (arctan (\frac{4}{9} x)) + \frac{1}{2} ln tan (arctan (\frac{4}{9} x)) + sec (arctan (\frac{4}{9} x))) : \frac{1}{128} (- 81 ln (\frac{4 x + 16 x ^{2} + 81}{9}) + 4 x 16 x^{2} + 81)

= \frac{1}{128} (- 81 ln (\frac{4 x + 16 x ^{2} + 81}{9}) + 4 x 16 x^{2} + 81)

Add a constant to the solution

= \frac{1}{128} (- 81 ln (\frac{4 x + 16 x ^{2} + 81}{9}) + 4 x 16 x^{2} + 81) + C

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What is the integral of (x^2)/(sqrt(16x^2+81)) ?
The integral of (x^2)/(sqrt(16x^2+81)) is 1/128 (-81ln((|4x+sqrt(16x^2+81)|)/9)+4xsqrt(16x^2+81))+C