What is the integral of sqrt(θ^2+1) ?

The integral of sqrt(θ^2+1) is 1/2 θsqrt(1+θ^2)+1/2 ln|θ+sqrt(1+θ^2)|+C

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integral of sqrt(θ^2+1)

\int θ^{2} + 1 d θ

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

Solution steps

\int θ^{2} + 1 d θ

Apply Trigonometric Substitution

= \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 (θ)

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

Simplify

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

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

Add a constant to the solution

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

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