What is the integral of |4x+1| ?

The integral of |4x+1| is 1/8 |4x+1|(4x+1)+C

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integral of |4x+1|

\int ∣ 4 x + 1 ∣ d x

\frac{1}{8} ∣ 4 x + 1 ∣ (4 x + 1) + C

Solution steps

\int ∣ 4 x + 1 ∣ d x

Apply u-substitution

: \int ∣ u ∣ \frac{1}{4} d u

= \int ∣ u ∣ \frac{1}{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 ∣ u ∣ d u

Use the common integral:

\int ∣ u ∣ d u = \frac{u ∣ u ∣}{2}

= \frac{1}{4} \cdot \frac{u ∣ u ∣}{2}

Substitute back

u = 4 x + 1

= \frac{1}{4} \cdot \frac{( 4 x + 1 ) ∣ 4 x + 1 ∣}{2}

Simplify

\frac{1}{4} \cdot \frac{( 4 x + 1 ) ∣ 4 x + 1 ∣}{2} : \frac{1}{8} ∣ 4 x + 1 ∣ (4 x + 1)

= \frac{1}{8} ∣ 4 x + 1 ∣ (4 x + 1)

Add a constant to the solution

= \frac{1}{8} ∣ 4 x + 1 ∣ (4 x + 1) + C

\int sin (e^{t}) d t

\frac{\partial}{\partial x} (\frac{4 x ^{2} y ^{2}}{x + y})

f (x) = x^{2} - 6 x + 5

\int_{- 4}^{3} \frac{1}{2} x d x

\frac{t}{1 + t ^{2}}