What is the integral of sqrt(1+(4x)^2) ?

The integral of sqrt(1+(4x)^2) is 1/8 (4xsqrt(16x^2+1)+ln|4x+sqrt(1+16x^2)|)+C

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integral of sqrt(1+(4x)^2)

\int 1 + (4 x)^{2} d x

\frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2}) + C

Solution steps

\int 1 + (4 x)^{2} d x

Simplify

1 + (4 x)^{2} : 1 + 16 x^{2}

= \int 1 + 16 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} sec^{3} (u) d u

= \int \frac{1}{4} 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}{4} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{4} (\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}{4} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (4 x)

= \frac{1}{4} (\frac{sec ^{2} ( arctan ( 4 x ) ) sin ( arctan ( 4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣)

Simplify

\frac{1}{4} (\frac{sec ^{2} ( arctan ( 4 x ) ) sin ( arctan ( 4 x ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (4 x)) + sec (arctan (4 x)) ∣) : \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2})

= \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2})

Add a constant to the solution

= \frac{1}{8} (4 x 16 x^{2} + 1 + ln 4 x + 1 + 16 x^{2}) + C

\frac{d}{d t} (2 t e^{- t})

\int \frac{1}{sin ( x ) cos ^{2} ( x )} d x

f (x) = 10 sin (x) + 103 cos (x)

\frac{2}{s ^{2} + 3 s + 2}

n = 5 \sum \infty \frac{n ^{2}}{n !}

What is the integral of sqrt(1+(4x)^2) ?
The integral of sqrt(1+(4x)^2) is 1/8 (4xsqrt(16x^2+1)+ln|4x+sqrt(1+16x^2)|)+C