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

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

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integral of sqrt(a^2+t^2)

\int a^{2} + t^{2} d t

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

Solution steps

\int a^{2} + t^{2} d t

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= a^{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) ∣

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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