What is the integral of (sqrt(x^2-49))/x ?

The integral of (sqrt(x^2-49))/x is -7arcsec(1/7 x)+sqrt(x^2-49)+C

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integral of (sqrt(x^2-49))/x

\int \frac{x ^{2} - 49}{x} d x

- 7 arcsec (\frac{1}{7} x) + x^{2} - 49 + C

Solution steps

\int \frac{x ^{2} - 49}{x} d x

Apply Trigonometric Substitution

: \int 7 tan^{2} (u) d u

= \int 7 tan^{2} (u) d u

Take the constant out:

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

= 7 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 7 \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

= 7 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 7 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{7} x)

= 7 (- arcsec (\frac{1}{7} x) + tan (arcsec (\frac{1}{7} x)))

Simplify

7 (- arcsec (\frac{1}{7} x) + tan (arcsec (\frac{1}{7} x))) : - 7 arcsec (\frac{1}{7} x) + x^{2} - 49

= - 7 arcsec (\frac{1}{7} x) + x^{2} - 49

Add a constant to the solution

= - 7 arcsec (\frac{1}{7} x) + x^{2} - 49 + C

x + y^{2} = 2, x + y = 0

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What is the integral of (sqrt(x^2-49))/x ?
The integral of (sqrt(x^2-49))/x is -7arcsec(1/7 x)+sqrt(x^2-49)+C