What is the integral of 1/(sqrt(256+x^2)) ?

The integral of 1/(sqrt(256+x^2)) is ln(1/16 |x+sqrt(256+x^2)|)+C

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integral of 1/(sqrt(256+x^2))

\int \frac{1}{256 + x ^{2}} d x

ln (\frac{1}{16} x + 256 + x^{2}) + C

Solution steps

\int \frac{1}{256 + x ^{2}} 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}{16} x)

= ln tan (arctan (\frac{1}{16} x)) + sec (arctan (\frac{1}{16} x))

Simplify

ln tan (arctan (\frac{1}{16} x)) + sec (arctan (\frac{1}{16} x)) : ln (\frac{1}{16} x + 256 + x^{2})

= ln (\frac{1}{16} x + 256 + x^{2})

Add a constant to the solution

= ln (\frac{1}{16} x + 256 + x^{2}) + C

e^{x c o s (2 x)}

\frac{d r}{d θ} = \frac{r ^{2} + 9 r θ ^{2}}{3 θ ^{3}}

\int_{0}^{1} 9 cos (3 x) d x

\int \frac{1}{( x + 3 ) x ^{2} + 6 x} d x

f (x) = \frac{x}{7} - 1

What is the integral of 1/(sqrt(256+x^2)) ?
The integral of 1/(sqrt(256+x^2)) is ln(1/16 |x+sqrt(256+x^2)|)+C