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)))

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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

= \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))

f (x) = 2 x^{3} - 6 x^{2} - 90 x - 1

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

\int \frac{1}{50} d x

2

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