What is the integral of sqrt(12+4x^2) ?

The integral of sqrt(12+4x^2) is xsqrt(3+x^2)+3ln| x/(sqrt(3))+sqrt(1/3 (x^2+3))|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(12+4x^2)

\int 12 + 4 x^{2} d x

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

Solution steps

\int 12 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int 6 sec^{3} (u) d u

= \int 6 sec^{3} (u) d u

Take the constant out:

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

= 6 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 6 (\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) ∣

= 6 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{2 3} x)

= 6 \frac{sec ^{2} ( arctan ( \frac{2}{2 3} x ) ) sin ( arctan ( \frac{2}{2 3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{2 3} x)) + sec (arctan (\frac{2}{2 3} x))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((\frac{1 + sin ( x )}{1 - cos ( x )})^{2})

\int_{- 3}^{2} (- x^{2} - x + 6) d x

\int (3 y^{2} + 6 y) (y^{3} + 3 y^{2} + 1)^{\frac{2}{3}} d y

f (x) = x^{2} - 8

\frac{d}{d x} (\frac{x + 3}{∣ x + 3 ∣})

What is the integral of sqrt(12+4x^2) ?
The integral of sqrt(12+4x^2) is xsqrt(3+x^2)+3ln| x/(sqrt(3))+sqrt(1/3 (x^2+3))|+C