What is the integral of sqrt(1+r^2) ?

The integral of sqrt(1+r^2) is 1/2 rsqrt(1+r^2)+1/2 ln|r+sqrt(1+r^2)|+C

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integral of sqrt(1+r^2)

\int 1 + r^{2} d r

\frac{1}{2} r 1 + r^{2} + \frac{1}{2} ln r + 1 + r^{2} + C

Solution steps

\int 1 + r^{2} d r

Apply Trigonometric Substitution

: \int sec^{3} (u) d u

= \int sec^{3} (u) d u

Apply Integral Reduction

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

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

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

Substitute back

u = arctan (r)

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

Simplify

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

= \frac{1}{2} r 1 + r^{2} + \frac{1}{2} ln r + 1 + r^{2}

Add a constant to the solution

= \frac{1}{2} r 1 + r^{2} + \frac{1}{2} ln r + 1 + r^{2} + C

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What is the integral of sqrt(1+r^2) ?
The integral of sqrt(1+r^2) is 1/2 rsqrt(1+r^2)+1/2 ln|r+sqrt(1+r^2)|+C