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 ^{2}}{d x ^{2}} (6 x^{2} + 5 x)

(t + 1) d t - \frac{1}{y ^{2}} d y = 0

\int \frac{2 w - 5}{w ^{2} - 3 w + \frac{9}{2}} d x

2 x + y^{2} = 8, x = y

\frac{\partial}{\partial x} (6 x y - 2 y^{2})

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