What is the integral of sqrt(1+16x^2) ?

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

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integral of sqrt(1+16x^2)

\int 1 + 16 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 + 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

(x + 1) y^{^{'}} + y = x

x \to - 3 lim (5)

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

x \to - 1 lim (\frac{4 x + 3}{2 x + 1})

\int 10 e^{\frac{x}{20}} d x

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