What is the integral of (1+x^2)^{1/2} ?

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

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integral of (1+x^2)^{1/2}

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

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

Solution steps

\int (1 + x^{2})^{\frac{1}{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 (x^{2} + 1)^{\frac{1}{2}} + \frac{1}{2} ln x + 1 + x^{2}

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

Add a constant to the solution

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

\int 2 (sin (2 x))^{2} d x

\frac{\partial}{\partial x} (x y \cdot e^{x y})

\frac{d}{d x} (3 x^{2} - e^{x})

\int 4^{t} sin (4^{t}) d t

\frac{d}{d x} (y - x + 1 + x - y + 1)

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