What is the integral of (sqrt(x^2-400))/x ?

The integral of (sqrt(x^2-400))/x is -20arcsec(1/20 x)+sqrt(x^2-400)+C

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integral of (sqrt(x^2-400))/x

\int \frac{x ^{2} - 400}{x} d x

- 20 arcsec (\frac{1}{20} x) + x^{2} - 400 + C

Solution steps

\int \frac{x ^{2} - 400}{x} d x

Apply Trigonometric Substitution

: \int 20 tan^{2} (u) d u

= \int 20 tan^{2} (u) d u

Take the constant out:

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

= 20 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 20 \cdot \int - 1 + sec^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 20 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= 20 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{20} x)

= 20 (- arcsec (\frac{1}{20} x) + tan (arcsec (\frac{1}{20} x)))

Simplify

20 (- arcsec (\frac{1}{20} x) + tan (arcsec (\frac{1}{20} x))) : - 20 arcsec (\frac{1}{20} x) + x^{2} - 400

= - 20 arcsec (\frac{1}{20} x) + x^{2} - 400

Add a constant to the solution

= - 20 arcsec (\frac{1}{20} x) + x^{2} - 400 + C

\int \frac{0}{x + 1} d x

x \to 0 lim (e^{2 x})

\frac{\partial}{\partial x} (- e^{y} sin (x) + 4 y + 7 y x)

\frac{r}{r ^{2} + 8}

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = 5

What is the integral of (sqrt(x^2-400))/x ?
The integral of (sqrt(x^2-400))/x is -20arcsec(1/20 x)+sqrt(x^2-400)+C