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

The integral of sqrt((1+(4x+2)^2)) is 1/8 (2(2x+1)sqrt(16x^2+16x+5)+ln|4x+2+sqrt(16x^2+16x+5)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{8} (2 (2 x + 1) 16 x^{2} + 16 x + 5 + ln 4 x + 2 + 16 x^{2} + 16 x + 5) + C

Solution steps

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

Apply u-substitution

: \int \frac{1 + u ^{2}}{4} d u

= \int \frac{1 + u ^{2}}{4} 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 1 + u^{2} d u

Apply Trigonometric Substitution

: \int sec^{3} (v) d v

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

Apply Integral Reduction

= \frac{1}{4} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

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

Substitute back

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

Simplify

\frac{1}{4} (\frac{sec ^{2} ( arctan ( 4 x + 2 ) ) sin ( arctan ( 4 x + 2 ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (4 x + 2)) + sec (arctan (4 x + 2)) ∣) : \frac{1}{8} (2 (2 x + 1) 16 x^{2} + 16 x + 5 + ln 4 x + 2 + 16 x^{2} + 16 x + 5)

= \frac{1}{8} (2 (2 x + 1) 16 x^{2} + 16 x + 5 + ln 4 x + 2 + 16 x^{2} + 16 x + 5)

Add a constant to the solution

= \frac{1}{8} (2 (2 x + 1) 16 x^{2} + 16 x + 5 + ln 4 x + 2 + 16 x^{2} + 16 x + 5) + C

\int \frac{5 x ^{2} + 2 x e ^{x} - 1}{x} d x

\int x^{3} sin (x) cos (x) d x

\int tan^{4} (x) \cdot sec^{4} (x) d x

\int 18 θ sin (18 θ^{2}) d θ

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

What is the integral of sqrt((1+(4x+2)^2)) ?
The integral of sqrt((1+(4x+2)^2)) is 1/8 (2(2x+1)sqrt(16x^2+16x+5)+ln|4x+2+sqrt(16x^2+16x+5)|)+C