What is the integral from 0 to 1 of sqrt(35+x^2) ?

The integral from 0 to 1 of sqrt(35+x^2) is 3+35/4 ln(7/5)

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integral from 0 to 1 of sqrt(35+x^2)

\int_{0}^{1} 35 + x^{2} d x

3 + \frac{35}{4} ln (\frac{7}{5})

Decimal Notation

5.94413 \dots

Solution steps

\int_{0}^{1} 35 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{35})} 35 sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{1}{35})} 35 sec^{3} (u) d u

Take the constant out:

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

= 35 \cdot \int_{0}^{a r c t a n (\frac{1}{35})} sec^{3} (u) d u

Apply Integral Reduction

= 35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{1}{35})} sec (u) d u

\int_{0}^{a r c t a n (\frac{1}{35})} sec (u) d u = \frac{1}{2} ln (\frac{7}{5})

= 35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \frac{1}{2} ln (\frac{7}{5})

Simplify

35 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{1}{35})} + \frac{1}{2} \cdot \frac{1}{2} ln (\frac{7}{5}) : 35 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{35})} + \frac{35}{4} ln (\frac{7}{5})

= 35 [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{1}{35})} + \frac{35}{4} ln (\frac{7}{5})

Compute the boundaries:

\frac{3}{35}

= 35 \cdot \frac{3}{35} + \frac{35}{4} ln (\frac{7}{5})

Simplify

= 3 + \frac{35}{4} ln (\frac{7}{5})

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What is the integral from 0 to 1 of sqrt(35+x^2) ?
The integral from 0 to 1 of sqrt(35+x^2) is 3+35/4 ln(7/5)