What is the integral of (x^2)/(3sqrt(x^2+4)) ?

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

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integral of (x^2)/(3sqrt(x^2+4))

\int \frac{x ^{2}}{3 x ^{2} + 4} d x

\frac{1}{6} (- 4 ln (\frac{x + x ^{2} + 4}{2}) + x x^{2} + 4) + C

Solution steps

\int \frac{x ^{2}}{3 x ^{2} + 4} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x ^{2}}{x ^{2} + 4} d x

Apply Trigonometric Substitution

: \int 4 tan^{2} (u) sec (u) d u

= \frac{1}{3} \cdot \int 4 tan^{2} (u) sec (u) d u

Take the constant out:

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

= \frac{1}{3} \cdot 4 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot 4 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \frac{1}{3} \cdot 4 \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{1}{3} \cdot 4 (- \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{1}{3} \cdot 4 (- 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{1}{2} x)

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

Simplify

\frac{1}{3} \cdot 4 (- ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x)) + \frac{1}{2} sec (arctan (\frac{1}{2} x)) tan (arctan (\frac{1}{2} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{2} x)) + sec (arctan (\frac{1}{2} x))) : \frac{1}{6} (- 4 ln (\frac{x + x ^{2} + 4}{2}) + x x^{2} + 4)

= \frac{1}{6} (- 4 ln (\frac{x + x ^{2} + 4}{2}) + x x^{2} + 4)

Add a constant to the solution

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

\int 4 x^{3} + 9 sin (x) d x

x \to 3 lim (x + 1)

\int sech^{2} (x) d x

n = 1 \sum \infty n! (x + 1)^{n}

n = 1 \sum \infty (- 1)^{n} \cdot \frac{1}{n}

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