What is the integral of sqrt(x^2+121) ?

The integral of sqrt(x^2+121) is 121(1/242 xsqrt(121+x^2)+1/2 ln((|x+sqrt(121+x^2)|)/(11)))+C

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integral of sqrt(x^2+121)

\int x^{2} + 121 d x

121 (\frac{1}{242} x 121 + x^{2} + \frac{1}{2} ln (\frac{x + 121 + x ^{2}}{11})) + C

Solution steps

\int x^{2} + 121 d x

Apply Trigonometric Substitution

: \int 121 sec^{3} (u) d u

= \int 121 sec^{3} (u) d u

Take the constant out:

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

= 121 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 121 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= 121 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{11} x)

= 121 (\frac{sec ^{2} ( arctan ( \frac{1}{11} x ) ) sin ( arctan ( \frac{1}{11} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{11} x)) + sec (arctan (\frac{1}{11} x)))

Simplify

121 (\frac{sec ^{2} ( arctan ( \frac{1}{11} x ) ) sin ( arctan ( \frac{1}{11} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{11} x)) + sec (arctan (\frac{1}{11} x))) : 121 (\frac{1}{242} x 121 + x^{2} + \frac{1}{2} ln (\frac{x + 121 + x ^{2}}{11}))

= 121 (\frac{1}{242} x 121 + x^{2} + \frac{1}{2} ln (\frac{x + 121 + x ^{2}}{11}))

Add a constant to the solution

= 121 (\frac{1}{242} x 121 + x^{2} + \frac{1}{2} ln (\frac{x + 121 + x ^{2}}{11})) + C

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What is the integral of sqrt(x^2+121) ?
The integral of sqrt(x^2+121) is 121(1/242 xsqrt(121+x^2)+1/2 ln((|x+sqrt(121+x^2)|)/(11)))+C