What is the integral of sqrt(49x^2+1764) ?

The integral of sqrt(49x^2+1764) is 252(1/72 xsqrt(x^2+36)+1/2 ln((|x+sqrt(36+x^2)|)/6))+C

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integral of sqrt(49x^2+1764)

\int 49 x^{2} + 1764 d x

252 (\frac{1}{72} x x^{2} + 36 + \frac{1}{2} ln (\frac{x + 36 + x ^{2}}{6})) + C

Solution steps

\int 49 x^{2} + 1764 d x

Apply Trigonometric Substitution

: \int 252 sec^{3} (u) d u

= \int 252 sec^{3} (u) d u

Take the constant out:

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

= 252 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= 252 (\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) ∣

= 252 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{7}{42} x)

= 252 (\frac{sec ^{2} ( arctan ( \frac{7}{42} x ) ) sin ( arctan ( \frac{7}{42} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{42} x)) + sec (arctan (\frac{7}{42} x)))

Simplify

252 (\frac{sec ^{2} ( arctan ( \frac{7}{42} x ) ) sin ( arctan ( \frac{7}{42} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{7}{42} x)) + sec (arctan (\frac{7}{42} x))) : 252 (\frac{1}{72} x x^{2} + 36 + \frac{1}{2} ln (\frac{x + 36 + x ^{2}}{6}))

= 252 (\frac{1}{72} x x^{2} + 36 + \frac{1}{2} ln (\frac{x + 36 + x ^{2}}{6}))

Add a constant to the solution

= 252 (\frac{1}{72} x x^{2} + 36 + \frac{1}{2} ln (\frac{x + 36 + x ^{2}}{6})) + C

\int \frac{x \cdot e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

\frac{\partial}{\partial y} (cos (x y^{2}))

x \to - 2 lim (x - 1)

\int sin (x) d sin (x) d x

x \to 3 lim (x^{3} - x^{2} + 15)

What is the integral of sqrt(49x^2+1764) ?
The integral of sqrt(49x^2+1764) is 252(1/72 xsqrt(x^2+36)+1/2 ln((|x+sqrt(36+x^2)|)/6))+C