What is the integral of sqrt(6+x^2) ?

The integral of sqrt(6+x^2) is 6(1/12 xsqrt(6+x^2)+1/2 ln| 1/(sqrt(6))x+sqrt(1/6 (6+x^2))|)+C

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integral of sqrt(6+x^2)

\int 6 + x^{2} d x

6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})) + C

Solution steps

\int 6 + x^{2} d x

Apply Trigonometric Substitution

= \int 6 sec^{3} (u) d u

Take the constant out:

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

= 6 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 6 (\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) ∣

= 6 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{6} x)

= 6 \frac{sec ^{2} ( arctan ( \frac{1}{6} x ) ) sin ( arctan ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x))

Simplify

6 \frac{sec ^{2} ( arctan ( \frac{1}{6} x ) ) sin ( arctan ( \frac{1}{6} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x)) : 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}))

= 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2}))

Add a constant to the solution

= 6 (\frac{1}{12} x 6 + x^{2} + \frac{1}{2} ln \frac{1}{6} x + \frac{1}{6} (6 + x^{2})) + C

x \to - 1 lim (x + 1)

\frac{d}{d x} ((ln (3 x))^{2})

d er i v a t i v e f (x) = \frac{2 x + x ^{2}}{( 1 + x ) ^{2}}

\frac{\partial}{\partial x} (\frac{Y ( x )}{x})

x y^{^{'}} - 4 y = 6 x^{4} e^{x}

What is the integral of sqrt(6+x^2) ?
The integral of sqrt(6+x^2) is 6(1/12 xsqrt(6+x^2)+1/2 ln| 1/(sqrt(6))x+sqrt(1/6 (6+x^2))|)+C