What is the integral of sqrt(x^2+8) ?

The integral of sqrt(x^2+8) is 8(1/16 xsqrt(8+x^2)+1/4 (2ln|x+sqrt(8+x^2)|-3ln(2)))+C

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integral of sqrt(x^2+8)

\int x^{2} + 8 d x

8 (\frac{1}{16} x 8 + x^{2} + \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2))) + C

Solution steps

\int x^{2} + 8 d x

Apply Trigonometric Substitution

: \int 8 sec^{3} (u) d u

= \int 8 sec^{3} (u) d u

Take the constant out:

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

= 8 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 8 (\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) ∣

= 8 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

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

= 8 \frac{sec ^{2} ( arctan ( \frac{1}{2 2} x ) ) sin ( arctan ( \frac{1}{2 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x))

Simplify

8 \frac{sec ^{2} ( arctan ( \frac{1}{2 2} x ) ) sin ( arctan ( \frac{1}{2 2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x)) : 8 (\frac{1}{16} x 8 + x^{2} + \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)))

= 8 (\frac{1}{16} x 8 + x^{2} + \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2)))

Add a constant to the solution

= 8 (\frac{1}{16} x 8 + x^{2} + \frac{1}{4} (2 ln x + 8 + x^{2} - 3 ln (2))) + C

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What is the integral of sqrt(x^2+8) ?
The integral of sqrt(x^2+8) is 8(1/16 xsqrt(8+x^2)+1/4 (2ln|x+sqrt(8+x^2)|-3ln(2)))+C