What is the integral from 0 to 1 of sqrt(2+5x^2) ?

The integral from 0 to 1 of sqrt(2+5x^2) is 1/(sqrt(5))((sqrt(35))/2+ln(sqrt(5/2)+sqrt(7/2)))

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integral from 0 to 1 of sqrt(2+5x^2)

\int_{0}^{1} 2 + 5 x^{2} d x

\frac{1}{5} (\frac{35}{2} + ln (\frac{5}{2} + \frac{7}{2}))

Decimal Notation

1.87694 \dots

Solution steps

\int_{0}^{1} 2 + 5 x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{5}{2})} \frac{2}{5} sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{5}{2})} \frac{2}{5} sec^{3} (u) d u

Take the constant out:

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

= \frac{2}{5} \cdot \int_{0}^{a r c t a n (\frac{5}{2})} sec^{3} (u) d u

Apply Integral Reduction

= \frac{2}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{5}{2})} sec (u) d u

\int_{0}^{a r c t a n (\frac{5}{2})} sec (u) d u = ln (\frac{5}{2} + \frac{7}{2})

= \frac{2}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (\frac{5}{2} + \frac{7}{2})

Simplify

\frac{2}{5} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + \frac{1}{2} ln (\frac{5}{2} + \frac{7}{2}) : \frac{1}{5} 2 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + ln (\frac{5}{2} + \frac{7}{2})

= \frac{1}{5} 2 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{5}{2})} + ln (\frac{5}{2} + \frac{7}{2})

Compute the boundaries:

\frac{35}{4}

= \frac{1}{5} (2 \cdot \frac{35}{4} + ln (\frac{5}{2} + \frac{7}{2}))

Simplify

= \frac{1}{5} (\frac{35}{2} + ln (\frac{5}{2} + \frac{7}{2}))

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

\frac{\partial}{\partial x} (4 + x + x^{2} Y (x))

\frac{d y}{d x} = x + 2

\int 2 x + 9 x^{- 2} d x

What is the integral from 0 to 1 of sqrt(2+5x^2) ?
The integral from 0 to 1 of sqrt(2+5x^2) is 1/(sqrt(5))((sqrt(35))/2+ln(sqrt(5/2)+sqrt(7/2)))