What is the integral of sqrt(x^2+17) ?

The integral of sqrt(x^2+17) is 17(1/34 xsqrt(17+x^2)+1/2 ln| 1/(sqrt(17))x+sqrt(1/17 (17+x^2))|)+C

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integral of sqrt(x^2+17)

\int x^{2} + 17 d x

17 (\frac{1}{34} x 17 + x^{2} + \frac{1}{2} ln \frac{1}{17} x + \frac{1}{17} (17 + x^{2})) + C

Solution steps

\int x^{2} + 17 d x

Apply Trigonometric Substitution

: \int 17 sec^{3} (u) d u

= \int 17 sec^{3} (u) d u

Take the constant out:

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

= 17 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 17 (\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) ∣

= 17 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{17} x)

= 17 \frac{sec ^{2} ( arctan ( \frac{1}{17} x ) ) sin ( arctan ( \frac{1}{17} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{17} x)) + sec (arctan (\frac{1}{17} x))

Simplify

17 \frac{sec ^{2} ( arctan ( \frac{1}{17} x ) ) sin ( arctan ( \frac{1}{17} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{17} x)) + sec (arctan (\frac{1}{17} x)) : 17 (\frac{1}{34} x 17 + x^{2} + \frac{1}{2} ln \frac{1}{17} x + \frac{1}{17} (17 + x^{2}))

= 17 (\frac{1}{34} x 17 + x^{2} + \frac{1}{2} ln \frac{1}{17} x + \frac{1}{17} (17 + x^{2}))

Add a constant to the solution

= 17 (\frac{1}{34} x 17 + x^{2} + \frac{1}{2} ln \frac{1}{17} x + \frac{1}{17} (17 + x^{2})) + C

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What is the integral of sqrt(x^2+17) ?
The integral of sqrt(x^2+17) is 17(1/34 xsqrt(17+x^2)+1/2 ln| 1/(sqrt(17))x+sqrt(1/17 (17+x^2))|)+C