What is the integral of (sqrt(a^2+u^2))/u ?

The integral of (sqrt(a^2+u^2))/u is u(1/2 ln((|a+sqrt(a^2+u^2)|)/(|u|))+(asqrt(a^2+u^2))/(2u^2))+C

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integral of (sqrt(a^2+u^2))/u

\int \frac{a ^{2} + u ^{2}}{u}

u (\frac{1}{2} ln (\frac{a + a ^{2} + u ^{2}}{∣ u ∣}) + \frac{a a ^{2} + u ^{2}}{2 u ^{2}}) + C

Solution steps

\int \frac{a ^{2} + u ^{2}}{u} d a

Take the constant out:

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

= \frac{1}{u} \cdot \int u^{2} + a^{2} d a

Apply Trigonometric Substitution

= \frac{1}{u} \cdot \int u^{2} sec^{3} (v) d v

Take the constant out:

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

= \frac{1}{u} u^{2} \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{u} u^{2} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{u} u^{2} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

v = arctan (\frac{1}{u} a)

= \frac{1}{u} u^{2} (\frac{sec ^{2} ( arctan ( \frac{1}{u} a ) ) sin ( arctan ( \frac{1}{u} a ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{u} a)) + sec (arctan (\frac{1}{u} a)))

Simplify

\frac{1}{u} u^{2} (\frac{sec ^{2} ( arctan ( \frac{1}{u} a ) ) sin ( arctan ( \frac{1}{u} a ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{u} a)) + sec (arctan (\frac{1}{u} a))) : u (\frac{1}{2} ln (\frac{a + a ^{2} + u ^{2}}{∣ u ∣}) + \frac{a a ^{2} + u ^{2}}{2 u ^{2}})

= u (\frac{1}{2} ln (\frac{a + a ^{2} + u ^{2}}{∣ u ∣}) + \frac{a a ^{2} + u ^{2}}{2 u ^{2}})

Add a constant to the solution

= u (\frac{1}{2} ln (\frac{a + a ^{2} + u ^{2}}{∣ u ∣}) + \frac{a a ^{2} + u ^{2}}{2 u ^{2}}) + C

What is the integral of (sqrt(a^2+u^2))/u ?
The integral of (sqrt(a^2+u^2))/u is u(1/2 ln((|a+sqrt(a^2+u^2)|)/(|u|))+(asqrt(a^2+u^2))/(2u^2))+C