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

f (x) = 3 x arccos (x)

\int \frac{1}{1} d x

x \to \infty lim (0.5)

\frac{d}{d x} (x (x - 5)^{2})

\frac{\partial}{\partial y} (- \frac{1}{x - y})

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