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

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

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integral of sqrt(x^2+49)

\int x^{2} + 49 d x

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

Solution steps

\int x^{2} + 49 d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

= 49 (\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) ∣

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

Substitute back

u = arctan (\frac{1}{7} x)

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

Simplify

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

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

Add a constant to the solution

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

y = 2 x, y = \frac{2 x}{3}

\frac{d}{d x} (2 x + y)^{3} = x

\frac{\partial}{\partial x} (cos^{3} (x))

\int \frac{x + 16}{x} d x

x \to - 3 lim (- (3 x))

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