What is the integral from 0 to 9 of sqrt(x^2+81) ?

The integral from 0 to 9 of sqrt(x^2+81) is (81(sqrt(2)+ln(1+sqrt(2))))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 9 of sqrt(x^2+81)

\int_{0}^{9} x^{2} + 81 d x

\frac{81 ( 2 + ln ( 1 + 2 ) )}{2}

Decimal Notation

92.97127 \dots

Solution steps

\int_{0}^{9} x^{2} + 81 d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 81 sec^{3} (u) d u

= \int_{0}^{\frac{π}{4}} 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_{0}^{\frac{π}{4}} sec^{3} (u) d u

Apply Integral Reduction

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (u) d u)

\int_{0}^{\frac{π}{4}} sec (u) d u = ln (1 + 2)

= 81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

81 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 81 ([\frac{1}{2} sec (u) tan (u)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 81 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= \frac{81 ( 2 + ln ( 1 + 2 ) )}{2}

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

y = ∣ 10 x ∣, y = x^{2} - 11

\frac{d x}{d t} = 2 \frac{x}{t} - 0.194 t

\frac{\partial}{\partial x} (ln (x^{6} - y^{6}))

\frac{d}{d M} (\frac{2 π r ^{\frac{3}{2}}}{GM})

What is the integral from 0 to 9 of sqrt(x^2+81) ?
The integral from 0 to 9 of sqrt(x^2+81) is (81(sqrt(2)+ln(1+sqrt(2))))/2