What is the integral of sqrt(1+100x^2) ?

The integral of sqrt(1+100x^2) is 1/20 (10xsqrt(100x^2+1)+ln|10x+sqrt(1+100x^2)|)+C

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integral of sqrt(1+100x^2)

\int 1 + 100 x^{2} d x

\frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2}) + C

Solution steps

\int 1 + 100 x^{2} d x

Apply Trigonometric Substitution

= \int \frac{1}{10} 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}{10} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{10} (\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}{10} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (10 x)

= \frac{1}{10} (\frac{sec ^{2} ( arctan ( 10 x ) ) sin ( arctan ( 10 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (10 x)) + sec (arctan (10 x)) ∣)

Simplify

\frac{1}{10} (\frac{sec ^{2} ( arctan ( 10 x ) ) sin ( arctan ( 10 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (10 x)) + sec (arctan (10 x)) ∣) : \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2})

= \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2})

Add a constant to the solution

= \frac{1}{20} (10 x 100 x^{2} + 1 + ln 10 x + 1 + 100 x^{2}) + C

\frac{1}{2} y^{^{'}} + 2 y = 0

\int_{x}^{\infty} 2 e^{- x - y} d y

d er i v a t i v e y = 4 x^{2} - 13 x - \frac{9}{x ^{4}}

t an g e n t f (x) = \frac{2}{x}, (- 3, - \frac{2}{3})

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

What is the integral of sqrt(1+100x^2) ?
The integral of sqrt(1+100x^2) is 1/20 (10xsqrt(100x^2+1)+ln|10x+sqrt(1+100x^2)|)+C