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Popular >

integral of sqrt(4x^2-16x+52)

Solution

\int 4 x^{2} - 16 x + 52 d x

Solution

x x^{2} - 4 x + 13 - 2 x^{2} - 4 x + 13 + 9 ln (\frac{1}{3} x + x^{2} - 4 x + 13 - 2) + C

Solution steps

\int 4 x^{2} - 16 x + 52 d x

Factor

4 x^{2} - 16 x + 52 : 2 x^{2} - 4 x + 13

= \int 2 x^{2} - 4 x + 13 d x

Take the constant out:

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

= 2 \cdot \int x^{2} - 4 x + 13 d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

= 2 \cdot \int (x - 2)^{2} + 9 d x

Apply u-substitution

: \int u^{2} + 9 d u

= 2 \cdot \int u^{2} + 9 d u

Apply Trigonometric Substitution

: \int 9 sec^{3} (v) d v

= 2 \cdot \int 9 sec^{3} (v) d v

Take the constant out:

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

= 2 \cdot 9 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 2 \cdot 9 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

= 2 \cdot 9 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 2 \cdot 9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{3} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2))))

Simplify

2 \cdot 9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{3} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2)))) : x x^{2} - 4 x + 13 - 2 x^{2} - 4 x + 13 + 9 ln (\frac{1}{3} x + x^{2} - 4 x + 13 - 2)

= x x^{2} - 4 x + 13 - 2 x^{2} - 4 x + 13 + 9 ln (\frac{1}{3} x + x^{2} - 4 x + 13 - 2)

Add a constant to the solution

= x x^{2} - 4 x + 13 - 2 x^{2} - 4 x + 13 + 9 ln (\frac{1}{3} x + x^{2} - 4 x + 13 - 2) + C

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