What is the integral of 1/(sqrt(100+x^2)) ?

The integral of 1/(sqrt(100+x^2)) is ln(1/10 |x+sqrt(100+x^2)|)+C

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integral of 1/(sqrt(100+x^2))

\int \frac{1}{100 + x ^{2}} d x

ln (\frac{1}{10} x + 100 + x^{2}) + C

Solution steps

\int \frac{1}{100 + 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}{10} x)

= ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x))

Simplify

ln tan (arctan (\frac{1}{10} x)) + sec (arctan (\frac{1}{10} x)) : ln (\frac{1}{10} x + 100 + x^{2})

= ln (\frac{1}{10} x + 100 + x^{2})

Add a constant to the solution

= ln (\frac{1}{10} x + 100 + x^{2}) + C

\frac{d}{d x} ((x + 1) (x - 1)^{2})

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

\frac{1}{x}

\frac{\partial}{\partial x} (5 y^{2} - x y + y^{3} sin (y))

\int (4 x - \frac{1}{2 x}) d x

What is the integral of 1/(sqrt(100+x^2)) ?
The integral of 1/(sqrt(100+x^2)) is ln(1/10 |x+sqrt(100+x^2)|)+C