What is the integral of 1/(sqrt(x^2+64)) ?

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

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integral of 1/(sqrt(x^2+64))

\int \frac{1}{x ^{2} + 64} d x

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

Solution steps

\int \frac{1}{x ^{2} + 64} d x

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x + 1}{e ^{x}})

f (x) = \frac{x ^{2} + 2 x - 5}{( x + 1 ) ^{2}}

t = \frac{1 - y ^{2}}{y ^{^{'}}}

y = 8

f (x) = \frac{1}{x ^{4}}, at x = - 3

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