What is the integral of sqrt(x^2+5) ?

The integral of sqrt(x^2+5) is 5(1/10 xsqrt(5+x^2)+1/2 ln| 1/(sqrt(5))x+sqrt(1/5 (5+x^2))|)+C

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integral of sqrt(x^2+5)

\int x^{2} + 5 d x

5 (\frac{1}{10} x 5 + x^{2} + \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (5 + x^{2})) + C

Solution steps

\int x^{2} + 5 d x

Apply Trigonometric Substitution

: \int 5 sec^{3} (u) d u

= \int 5 sec^{3} (u) d u

Take the constant out:

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

= 5 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 5 (\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) ∣

= 5 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{5} x)

= 5 \frac{sec ^{2} ( arctan ( \frac{1}{5} x ) ) sin ( arctan ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x))

Simplify

5 \frac{sec ^{2} ( arctan ( \frac{1}{5} x ) ) sin ( arctan ( \frac{1}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x)) : 5 (\frac{1}{10} x 5 + x^{2} + \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (5 + x^{2}))

= 5 (\frac{1}{10} x 5 + x^{2} + \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (5 + x^{2}))

Add a constant to the solution

= 5 (\frac{1}{10} x 5 + x^{2} + \frac{1}{2} ln \frac{1}{5} x + \frac{1}{5} (5 + x^{2})) + C

r \to 0 lim (sin^{2} (r))

(2 - 3 x^{3})^{- 2}

\frac{e ^{- 5 s}}{( s - 2 ) ^{4}}

\int x^{2} + 3 x + 1 d x

\int 2 y d y

What is the integral of sqrt(x^2+5) ?
The integral of sqrt(x^2+5) is 5(1/10 xsqrt(5+x^2)+1/2 ln| 1/(sqrt(5))x+sqrt(1/5 (5+x^2))|)+C