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

The integral from-1 to 1 of sqrt(1+4x^2) is sqrt(5)+1/4 ln(2+sqrt(5))-1/4 ln(-2+sqrt(5))

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

\int_{- 1}^{1} 1 + 4 x^{2} d x

5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

Decimal Notation

2.95788 \dots

Solution steps

\int_{- 1}^{1} 1 + 4 x^{2} d x

Apply Trigonometric Substitution

: \int_{- a r c t a n (2)}^{a r c t a n (2)} \frac{1}{2} sec^{3} (u) d u

= \int_{- a r c t a n (2)}^{a r c t a n (2)} \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_{- a r c t a n (2)}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

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

\int_{- a r c t a n (2)}^{a r c t a n (2)} sec (u) d u = ln (2 + 5) - ln (- 2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (2)}^{a r c t a n (2)} + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Compute the boundaries:

25

= \frac{1}{2} (25 + \frac{1}{2} (ln (2 + 5) - ln (- 2 + 5)))

Simplify

= 5 + \frac{1}{4} ln (2 + 5) - \frac{1}{4} ln (- 2 + 5)

\int (x + \frac{3}{x}) d x

x \to - 1 lim ((1 + x)^{- 1})

\frac{\partial}{\partial x} (\frac{3 x - y}{( x + y ) ^{3}})

y^{^{'}} = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}

\frac{\partial}{\partial x} (cos^{2} (x + y))

What is the integral from-1 to 1 of sqrt(1+4x^2) ?
The integral from-1 to 1 of sqrt(1+4x^2) is sqrt(5)+1/4 ln(2+sqrt(5))-1/4 ln(-2+sqrt(5))