What is the integral of sqrt(x^2+64) ?

The integral of sqrt(x^2+64) is 64(1/128 xsqrt(64+x^2)+1/2 ln((|x+sqrt(64+x^2)|)/8))+C

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integral of sqrt(x^2+64)

\int x^{2} + 64 d x

64 (\frac{1}{128} x 64 + x^{2} + \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8})) + C

Solution steps

\int x^{2} + 64 d x

Apply Trigonometric Substitution

: \int 64 sec^{3} (u) d u

= \int 64 sec^{3} (u) d u

Take the constant out:

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

= 64 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 64 (\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) ∣

= 64 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{8} x)

= 64 (\frac{sec ^{2} ( arctan ( \frac{1}{8} x ) ) sin ( arctan ( \frac{1}{8} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x)))

Simplify

64 (\frac{sec ^{2} ( arctan ( \frac{1}{8} x ) ) sin ( arctan ( \frac{1}{8} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{8} x)) + sec (arctan (\frac{1}{8} x))) : 64 (\frac{1}{128} x 64 + x^{2} + \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}))

= 64 (\frac{1}{128} x 64 + x^{2} + \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8}))

Add a constant to the solution

= 64 (\frac{1}{128} x 64 + x^{2} + \frac{1}{2} ln (\frac{x + 64 + x ^{2}}{8})) + C

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What is the integral of sqrt(x^2+64) ?
The integral of sqrt(x^2+64) is 64(1/128 xsqrt(64+x^2)+1/2 ln((|x+sqrt(64+x^2)|)/8))+C