What is the integral of x^2*1/(sqrt(4x^2+9)) ?

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

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integral of x^2*1/(sqrt(4x^2+9))

\int x^{2} \cdot \frac{1}{4 x ^{2} + 9} d x

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

Solution steps

\int x^{2} \frac{1}{4 x ^{2} + 9} d x

Simplify

x^{2} \frac{1}{4 x ^{2} + 9} : \frac{x ^{2}}{4 x ^{2} + 9}

= \int \frac{x ^{2}}{4 x ^{2} + 9} d x

Apply Trigonometric Substitution

: \int \frac{9 tan ^{2} ( u ) sec ( u )}{8} d u

= \int \frac{9 tan ^{2} ( u ) sec ( u )}{8} d u

Take the constant out:

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

= \frac{9}{8} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{9}{8} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \frac{9}{8} \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{9}{8} (- \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{9}{8} (- 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{2}{3} x)

= \frac{9}{8} (- ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x)) + \frac{1}{2} sec (arctan (\frac{2}{3} x)) tan (arctan (\frac{2}{3} x)) + \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x)))

Simplify

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

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

Add a constant to the solution

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

x^{2} + x + 1

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

\int \frac{1}{x - 1} d x

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

\int (\frac{x ^{2} - 3 x + 5}{x}) d x

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