What is the integral of (sqrt(x^2-441))/x ?

The integral of (sqrt(x^2-441))/x is -21arcsec(1/21 x)+sqrt(x^2-441)+C

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integral of (sqrt(x^2-441))/x

\int \frac{x ^{2} - 441}{x} d x

- 21 arcsec (\frac{1}{21} x) + x^{2} - 441 + C

Solution steps

\int \frac{x ^{2} - 441}{x} d x

Apply Trigonometric Substitution

: \int 21 tan^{2} (u) d u

= \int 21 tan^{2} (u) d u

Take the constant out:

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

= 21 \cdot \int tan^{2} (u) d u

Rewrite using trig identities

= 21 \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

= 21 (- \int 1 d u + \int sec^{2} (u) d u)

\int 1 d u = u

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

= 21 (- u + tan (u))

Substitute back

u = arcsec (\frac{1}{21} x)

= 21 (- arcsec (\frac{1}{21} x) + tan (arcsec (\frac{1}{21} x)))

Simplify

21 (- arcsec (\frac{1}{21} x) + tan (arcsec (\frac{1}{21} x))) : - 21 arcsec (\frac{1}{21} x) + x^{2} - 441

= - 21 arcsec (\frac{1}{21} x) + x^{2} - 441

Add a constant to the solution

= - 21 arcsec (\frac{1}{21} x) + x^{2} - 441 + C

lo g_{2} (3 x^{2} - 5)

\frac{s}{2 s + 1}

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}, y (1) = 1

\int_{- 2}^{2} 2 - 4 - x^{2} d x

\frac{1}{s + 5}

What is the integral of (sqrt(x^2-441))/x ?
The integral of (sqrt(x^2-441))/x is -21arcsec(1/21 x)+sqrt(x^2-441)+C