What is the integral of (29x^2)/(sqrt(x^2-4)) ?

The integral of (29x^2)/(sqrt(x^2-4)) is 116(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (29x^2)/(sqrt(x^2-4))

\int \frac{29 x ^{2}}{x ^{2} - 4} d x

116 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

Solution steps

\int \frac{29 x ^{2}}{x ^{2} - 4} d x

Take the constant out:

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

= 29 \cdot \int \frac{x ^{2}}{x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int 4 sec^{3} (u) d u

= 29 \cdot \int 4 sec^{3} (u) d u

Take the constant out:

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

= 29 \cdot 4 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 29 \cdot 4 (\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) ∣

= 29 \cdot 4 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{1}{2} x)

= 29 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x)))

Simplify

29 \cdot 4 (\frac{sec ^{2} ( arcsec ( \frac{1}{2} x ) ) sin ( arcsec ( \frac{1}{2} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{2} x)) + sec (arcsec (\frac{1}{2} x))) : 116 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

= 116 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2}))

Add a constant to the solution

= 116 (\frac{1}{8} x x^{2} - 4 + \frac{1}{2} ln (\frac{x ^{2} - 4 + x}{2})) + C

y = x^{2} + 2 x, y = x + 2

\int \frac{1}{8} t d t

y = tan (2 x)

\int \frac{( x )}{( 4 x ^{(2)} - 12 x + 25 )} d x

x \to 0 + lim (\frac{1}{1 - cos ( x )})

What is the integral of (29x^2)/(sqrt(x^2-4)) ?
The integral of (29x^2)/(sqrt(x^2-4)) is 116(1/8 xsqrt(x^2-4)+1/2 ln((|sqrt(x^2-4)+x|)/2))+C