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

The integral of (x^2)/(sqrt(9x^2+16)) is 1/54 (-16ln((|3x+sqrt(9x^2+16)|)/4)+3xsqrt(9x^2+16))+C

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integral of (x^2)/(sqrt(9x^2+16))

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

\frac{1}{54} (- 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}) + 3 x 9 x^{2} + 16) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{16 tan ^{2} ( u ) sec ( u )}{27} d u

= \int \frac{16 tan ^{2} ( u ) sec ( u )}{27} d u

Take the constant out:

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

= \frac{16}{27} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{16}{27} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \frac{16}{27} \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{16}{27} (- \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{16}{27} (- 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{3}{4} x)

= \frac{16}{27} (- ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x)) + \frac{1}{2} sec (arctan (\frac{3}{4} x)) tan (arctan (\frac{3}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x)))

Simplify

\frac{16}{27} (- ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x)) + \frac{1}{2} sec (arctan (\frac{3}{4} x)) tan (arctan (\frac{3}{4} x)) + \frac{1}{2} ln tan (arctan (\frac{3}{4} x)) + sec (arctan (\frac{3}{4} x))) : \frac{1}{54} (- 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}) + 3 x 9 x^{2} + 16)

= \frac{1}{54} (- 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}) + 3 x 9 x^{2} + 16)

Add a constant to the solution

= \frac{1}{54} (- 16 ln (\frac{3 x + 9 x ^{2} + 16}{4}) + 3 x 9 x^{2} + 16) + C

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What is the integral of (x^2)/(sqrt(9x^2+16)) ?
The integral of (x^2)/(sqrt(9x^2+16)) is 1/54 (-16ln((|3x+sqrt(9x^2+16)|)/4)+3xsqrt(9x^2+16))+C