What is the integral of sqrt(x^2+2x+50) ?

The integral of sqrt(x^2+2x+50) is 49(1/98 (x+1)sqrt(x^2+2x+50)+1/2 ln((|x+1+sqrt(x^2+2x+50)|)/7))+C

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integral of sqrt(x^2+2x+50)

\int x^{2} + 2 x + 50 d x

49 (\frac{1}{98} (x + 1) x^{2} + 2 x + 50 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 50}{7})) + C

Solution steps

\int x^{2} + 2 x + 50 d x

Complete the square

x^{2} + 2 x + 50 : (x + 1)^{2} + 49

= \int (x + 1)^{2} + 49 d x

Apply u-substitution

: \int u^{2} + 49 d u

= \int u^{2} + 49 d u

Apply Trigonometric Substitution

: \int 49 sec^{3} (v) d v

= \int 49 sec^{3} (v) d v

Take the constant out:

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

= 49 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 49 (\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) ∣

= 49 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 49 (\frac{sec ^{2} ( arctan ( \frac{1}{7} ( x + 1 ) ) ) sin ( arctan ( \frac{1}{7} ( x + 1 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} (x + 1))) + sec (arctan (\frac{1}{7} (x + 1))))

Simplify

49 (\frac{sec ^{2} ( arctan ( \frac{1}{7} ( x + 1 ) ) ) sin ( arctan ( \frac{1}{7} ( x + 1 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{7} (x + 1))) + sec (arctan (\frac{1}{7} (x + 1)))) : 49 (\frac{1}{98} (x + 1) x^{2} + 2 x + 50 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 50}{7}))

= 49 (\frac{1}{98} (x + 1) x^{2} + 2 x + 50 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 50}{7}))

Add a constant to the solution

= 49 (\frac{1}{98} (x + 1) x^{2} + 2 x + 50 + \frac{1}{2} ln (\frac{x + 1 + x ^{2} + 2 x + 50}{7})) + C

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What is the integral of sqrt(x^2+2x+50) ?
The integral of sqrt(x^2+2x+50) is 49(1/98 (x+1)sqrt(x^2+2x+50)+1/2 ln((|x+1+sqrt(x^2+2x+50)|)/7))+C