What is the integral from-1 to 1 of 1/(5-4x) ?

The integral from-1 to 1 of 1/(5-4x) is 1/2 ln(3)

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integral from-1 to 1 of 1/(5-4x)

\int_{- 1}^{1} \frac{1}{5 - 4 x} d x

\frac{1}{2} ln (3)

Decimal Notation

0.54930 \dots

Solution steps

\int_{- 1}^{1} \frac{1}{5 - 4 x} d x

Apply u-substitution

: \int_{9}^{1} - \frac{1}{4 u} d u

= \int_{9}^{1} - \frac{1}{4 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{9} - \frac{1}{4 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_{1}^{9} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{4} [ln ∣ u ∣]_{1}^{9})

Simplify

= \frac{1}{4} [ln ∣ u ∣]_{1}^{9}

Compute the boundaries:

2 ln (3)

= \frac{1}{4} \cdot 2 ln (3)

Simplify

= \frac{1}{2} ln (3)

\int_{2}^{4} 2 d x

\int_{1}^{2} (1 + \frac{1}{u} + \frac{1}{u ^{2}}) d u

\int_{- 2}^{0} 2 x^{2} e^{3 x^{3} + 7} d x

\int_{2}^{3} x^{4} d x

\int_{0}^{1} 3 y d y

What is the integral from-1 to 1 of 1/(5-4x) ?
The integral from-1 to 1 of 1/(5-4x) is 1/2 ln(3)