What is the integral of sqrt(1+36x^2) ?

The integral of sqrt(1+36x^2) is 1/12 (6xsqrt(36x^2+1)+ln|6x+sqrt(1+36x^2)|)+C

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integral of sqrt(1+36x^2)

\int 1 + 36 x^{2} d x

\frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2}) + C

Solution steps

\int 1 + 36 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} sec^{3} (u) d u

= \int \frac{1}{6} sec^{3} (u) d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{6} (\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) ∣

= \frac{1}{6} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (6 x)

= \frac{1}{6} (\frac{sec ^{2} ( arctan ( 6 x ) ) sin ( arctan ( 6 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (6 x)) + sec (arctan (6 x)) ∣)

Simplify

\frac{1}{6} (\frac{sec ^{2} ( arctan ( 6 x ) ) sin ( arctan ( 6 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (6 x)) + sec (arctan (6 x)) ∣) : \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2})

= \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2})

Add a constant to the solution

= \frac{1}{12} (6 x 36 x^{2} + 1 + ln 6 x + 1 + 36 x^{2}) + C

\int (x \frac{e ^{- 2 x}}{2}) d x

y = sin (x) + tan (x)

cos ((π x)^{3})

\frac{d}{d x} (ln (3 x^{2} - 2 x + 1))

\int_{0}^{π} cos^{3} (x) d x

What is the integral of sqrt(1+36x^2) ?
The integral of sqrt(1+36x^2) is 1/12 (6xsqrt(36x^2+1)+ln|6x+sqrt(1+36x^2)|)+C