What is the integral of (1+t^2)^{1/2} ?

The integral of (1+t^2)^{1/2} is 1/2 t(t^2+1)^{1/2}+1/2 ln|t+sqrt(1+t^2)|+C

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integral of (1+t^2)^{1/2}

\int (1 + t^{2})^{\frac{1}{2}} d t

\frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2} + C

Solution steps

\int (1 + t^{2})^{\frac{1}{2}} d t

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 (t)

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

Simplify

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

= \frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2}

Add a constant to the solution

= \frac{1}{2} t (t^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln t + 1 + t^{2} + C

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What is the integral of (1+t^2)^{1/2} ?
The integral of (1+t^2)^{1/2} is 1/2 t(t^2+1)^{1/2}+1/2 ln|t+sqrt(1+t^2)|+C