What is the integral from 0 to 3 of sqrt(1+4x^2) ?

The integral from 0 to 3 of sqrt(1+4x^2) is 1/2 (3sqrt(37)+1/2 ln(6+sqrt(37)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 3 of sqrt(1+4x^2)

\int_{0}^{3} 1 + 4 x^{2} d x

\frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

Decimal

9.74708 \dots

Solution steps

\int_{0}^{3} 1 + 4 x^{2} d x

Apply Trigonometric Substitution

= \int_{0}^{a r c t a n (6)} \frac{1}{2} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec (u) d u)

\int_{0}^{a r c t a n (6)} sec (u) d u = ln (6 + 37)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

Compute the boundaries:

337

= \frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

\frac{d}{d t} (a t^{2} + b t + c)

d er i v a t i v e 2

\int x sec^{4} (x^{2} + 1) tan^{7} (x^{2} + 1) d x

x \to 3 lim (\frac{5}{x ( x ^{3} - 27 )})

\int \frac{8 x}{x ^{3} + 1} d x

What is the integral from 0 to 3 of sqrt(1+4x^2) ?
The integral from 0 to 3 of sqrt(1+4x^2) is 1/2 (3sqrt(37)+1/2 ln(6+sqrt(37)))