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

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

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integral of sqrt(2+x^2)

\int 2 + x^{2} d x

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

Solution steps

\int 2 + x^{2} d x

Apply Trigonometric Substitution

: \int 2 sec^{3} (u) d u

= \int 2 sec^{3} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 2 (\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) ∣

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

Substitute back

u = arctan (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

x^{3} - 10 x^{2} + 12 x + 3

\int \frac{3 x ^{4}}{2} - 4 x^{\frac{3}{2}} d x

f (x) = \frac{e ^{x} + e ^{- x}}{2}

\frac{d y}{d x} = \frac{5 x + 2 y}{5 x + 2 y + 2}

\frac{d ^{2} y}{d x ^{2}} = 0

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