What is the integral of x/(1+4x) ?

The integral of x/(1+4x) is 1/16 (1+4x-ln|1+4x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(1+4x)

\int \frac{x}{1 + 4 x} d x

\frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

Solution steps

\int \frac{x}{1 + 4 x} d x

Apply u-substitution

: \int \frac{u - 1}{16 u} d u

= \int \frac{u - 1}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{16} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{16} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{16} (u - ln ∣ u ∣)

Substitute back

u = 1 + 4 x

= \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣)

Add a constant to the solution

= \frac{1}{16} (1 + 4 x - ln ∣ 1 + 4 x ∣) + C

\int \frac{1}{x ( 3 + x )} d x

\int t sin (t) cos (2 t) d t

\int 18 (x^{2} + 1) (x^{3} + 3 x)^{2} d x

\int \frac{5}{2 x} sin (2 x) d x

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

What is the integral of x/(1+4x) ?
The integral of x/(1+4x) is 1/16 (1+4x-ln|1+4x|)+C