What is the integral of sqrt(x^2+81) ?

The integral of sqrt(x^2+81) is 81(1/162 xsqrt(81+x^2)+1/2 ln((|x+sqrt(81+x^2)|)/9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(x^2+81)

\int x^{2} + 81 d x

81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9})) + C

Solution steps

\int x^{2} + 81 d x

Apply Trigonometric Substitution

: \int 81 sec^{3} (u) d u

= \int 81 sec^{3} (u) d u

Take the constant out:

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

= 81 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 81 (\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) ∣

= 81 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{9} x)

= 81 (\frac{sec ^{2} ( arctan ( \frac{1}{9} x ) ) sin ( arctan ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x)))

Simplify

81 (\frac{sec ^{2} ( arctan ( \frac{1}{9} x ) ) sin ( arctan ( \frac{1}{9} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x))) : 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9}))

= 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9}))

Add a constant to the solution

= 81 (\frac{1}{162} x 81 + x^{2} + \frac{1}{2} ln (\frac{x + 81 + x ^{2}}{9})) + C

x \to 0 lim (3 x)

y = 2 x^{2} + 7 x, (- 1, - 5)

\frac{d y}{d t} = 2 y + sin (2 t)

\frac{d y}{d x} = \frac{2 x sec ( \frac{y}{x} ) + y}{x}

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

What is the integral of sqrt(x^2+81) ?
The integral of sqrt(x^2+81) is 81(1/162 xsqrt(81+x^2)+1/2 ln((|x+sqrt(81+x^2)|)/9))+C