What is the integral from 1 to 3 of sqrt(u^2+b^2) ?

The integral from 1 to 3 of sqrt(u^2+b^2) is

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integral from 1 to 3 of sqrt(u^2+b^2)

\int_{1}^{3} u^{2} + b^{2} d u

b^{2} (\frac{3 b ^{2} + 9 - b ^{2} + 1}{2 b ^{2}} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})))

Solution steps

\int_{1}^{3} u^{2} + b^{2} d u

Apply Trigonometric Substitution

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

= \int_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} b^{2} sec^{3} (v) d v

Take the constant out:

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

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

Apply Integral Reduction

= b^{2} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} + \frac{1}{2} \cdot \int_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} sec (v) d v

\int_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} sec (v) d v = ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})

= b^{2} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣}))

Simplify

b^{2} [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})) : b^{2} ([\frac{1}{2} sec (v) tan (v)]_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})))

= b^{2} ([\frac{1}{2} sec (v) tan (v)]_{a r c t a n (\frac{1}{b})}^{a r c t a n (\frac{3}{b})} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})))

Compute the boundaries:

\frac{3 b ^{2} + 9}{2 b ^{2}} - \frac{b ^{2} + 1}{2 b ^{2}}

= b^{2} (\frac{3 b ^{2} + 9}{2 b ^{2}} - \frac{b ^{2} + 1}{2 b ^{2}} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})))

Simplify

= b^{2} (\frac{3 b ^{2} + 9 - b ^{2} + 1}{2 b ^{2}} + \frac{1}{2} (ln (\frac{3 + b ^{2} + 9}{∣ b ∣}) - ln (\frac{1 + b ^{2} + 1}{∣ b ∣})))

\int x - 2 π d x

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\frac{\partial}{\partial x} (arctan (x^{3} y^{2}))

\int (x) \frac{3}{4} x^{2} - 4 x + 8 d x

\int \frac{3}{x ^{\frac{2}{3}}} d x

What is the integral from 1 to 3 of sqrt(u^2+b^2) ?
The integral from 1 to 3 of sqrt(u^2+b^2) is