What is the integral of sqrt(7+x^2) ?

The integral of sqrt(7+x^2) is 7(1/14 xsqrt(7+x^2)+1/2 ln| 1/(sqrt(7))x+sqrt(1/7 (7+x^2))|)+C

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integral of sqrt(7+x^2)

\int 7 + x^{2} d x

7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2})) + C

Solution steps

\int 7 + x^{2} d x

Apply Trigonometric Substitution

: \int 7 sec^{3} (u) d u

= \int 7 sec^{3} (u) d u

Take the constant out:

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

= 7 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 7 (\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) ∣

= 7 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{7} x)

= 7 \frac{sec ^{2} ( arctan ( \frac{1}{7} x ) ) sin ( arctan ( \frac{1}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} x)) + sec (arctan (\frac{1}{7} x))

Simplify

7 \frac{sec ^{2} ( arctan ( \frac{1}{7} x ) ) sin ( arctan ( \frac{1}{7} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} x)) + sec (arctan (\frac{1}{7} x)) : 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2}))

= 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2}))

Add a constant to the solution

= 7 (\frac{1}{14} x 7 + x^{2} + \frac{1}{2} ln \frac{1}{7} x + \frac{1}{7} (7 + x^{2})) + C

x^{^{''}} - 0.1 x^{^{'}} + x = 0

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

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

x \to 0 - lim (F (x))

f (x) = 1 + 4 - x, at x = - 2

What is the integral of sqrt(7+x^2) ?
The integral of sqrt(7+x^2) is 7(1/14 xsqrt(7+x^2)+1/2 ln| 1/(sqrt(7))x+sqrt(1/7 (7+x^2))|)+C