What is the integral of sqrt(49x^2-441) ?

The integral of sqrt(49x^2-441) is 63(1/18 xsqrt(x^2-9)-1/2 ln((|x+sqrt(x^2-9)|)/3))+C

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integral of sqrt(49x^2-441)

\int 49 x^{2} - 441 d x

63 (\frac{1}{18} x x^{2} - 9 - \frac{1}{2} ln (\frac{x + x ^{2} - 9}{3})) + C

Solution steps

\int 49 x^{2} - 441 d x

Apply Trigonometric Substitution

: \int 63 tan^{2} (u) sec (u) d u

= \int 63 tan^{2} (u) sec (u) d u

Take the constant out:

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

= 63 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 63 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 63 \cdot \int - sec (u) + sec^{3} (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

= 63 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 63 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{7}{21} x)

= 63 (- ln tan (arcsec (\frac{7}{21} x)) + sec (arcsec (\frac{7}{21} x)) + \frac{1}{2} sec (arcsec (\frac{7}{21} x)) tan (arcsec (\frac{7}{21} x)) + \frac{1}{2} ln tan (arcsec (\frac{7}{21} x)) + sec (arcsec (\frac{7}{21} x)))

Simplify

63 (- ln tan (arcsec (\frac{7}{21} x)) + sec (arcsec (\frac{7}{21} x)) + \frac{1}{2} sec (arcsec (\frac{7}{21} x)) tan (arcsec (\frac{7}{21} x)) + \frac{1}{2} ln tan (arcsec (\frac{7}{21} x)) + sec (arcsec (\frac{7}{21} x))) : 63 (\frac{1}{18} x x^{2} - 9 - \frac{1}{2} ln (\frac{x + x ^{2} - 9}{3}))

= 63 (\frac{1}{18} x x^{2} - 9 - \frac{1}{2} ln (\frac{x + x ^{2} - 9}{3}))

Add a constant to the solution

= 63 (\frac{1}{18} x x^{2} - 9 - \frac{1}{2} ln (\frac{x + x ^{2} - 9}{3})) + C

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What is the integral of sqrt(49x^2-441) ?
The integral of sqrt(49x^2-441) is 63(1/18 xsqrt(x^2-9)-1/2 ln((|x+sqrt(x^2-9)|)/3))+C