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

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

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integral of sqrt(1+t^2)

\int 1 + t^{2} d t

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

Solution steps

\int 1 + t^{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 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

x \to - 1 lim (\frac{1}{x ^{2}})

- x^{3} (- 2.8)

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

\frac{d}{d θ} (7 sin (θ))

- 2 x sin (x^{2})

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