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

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

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integral of sqrt(2+t^2)

\int 2 + t^{2} d t

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

Solution steps

\int 2 + t^{2} d t

Apply Trigonometric Substitution

: \int 2 sec^{3} (u) d u

= \int 2 sec^{3} (u) d u

Take the constant out:

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

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

Apply Integral Reduction

= 2 (\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) ∣

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

Substitute back

u = arctan (\frac{1}{2} t)

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

Simplify

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

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

Add a constant to the solution

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

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