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

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

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integral of sqrt(t^2+1)

\int t^{2} + 1 d t

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

Solution steps

\int t^{2} + 1 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 1 + t^{2} + \frac{1}{2} ln t + 1 + t^{2}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2} - 1}{( x + 1 ) ^{2}})

\frac{d}{d t} (8)

\int \frac{e ^{\frac{7}{x}}}{x ^{2}} d x

x \to 0 lim (\frac{x - ∣ x ∣}{x})

\int \frac{1}{( x - 3 ) ln ( x - 3 )} d x

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