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

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

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integral of sqrt(1+x^2)

\int 1 + x^{2} d x

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

Solution steps

\int 1 + x^{2} 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{\partial}{\partial z} (\frac{1}{x y z ^{2}})

\frac{d y}{d x} = \frac{- x y ^{2}}{2}

x \to 2 lim ((x - 2) (2^{x} - 4))

x^{3^{x}}

\int \frac{( x + 2 ) ^{2}}{x ^{2} + 4} d x

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