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

The integral of sqrt(x^2+1) is 1/2 xsqrt(1+x^2)+1/2 ln|x+sqrt(1+x^2)|+C

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integral of sqrt(x^2+1)

\int x^{2} + 1 d x

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

Solution steps

\int x^{2} + 1 d x

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 (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{2 x - x^{2}})

n = 0 \sum \infty \frac{6 ^{n}}{n + 1}

x \to - \frac{7 π}{2} + lim (sec (x))

\int (x^{2} + 7) \cdot ln (x) d x

x \to 0 lim (\frac{1}{cos ( 19 x )})

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