What is the integral from 2 to 7 of 4/(8x-11) ?

The integral from 2 to 7 of 4/(8x-11) is ln(3)

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integral from 2 to 7 of 4/(8x-11)

\int_{2}^{7} \frac{4}{8 x - 11} d x

ln (3)

Solution steps

\int_{2}^{7} \frac{4}{8 x - 11} d x

Take the constant out:

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

= 4 \cdot \int_{2}^{7} \frac{1}{8 x - 11} d x

Apply u-substitution

: \int_{5}^{45} \frac{1}{8 u} d u

= 4 \cdot \int_{5}^{45} \frac{1}{8 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{8} \cdot \int_{5}^{45} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{8} [ln ∣ u ∣]_{5}^{45}

Simplify

4 \cdot \frac{1}{8} [ln ∣ u ∣]_{5}^{45} : \frac{1}{2} [ln ∣ u ∣]_{5}^{45}

= \frac{1}{2} [ln ∣ u ∣]_{5}^{45}

Compute the boundaries:

ln (45) - ln (5)

= \frac{1}{2} (ln (45) - ln (5))

Simplify

= ln (3)

\frac{6 x ^{2} + 2 x + 6}{x}

y = tan (x), at x = \frac{π}{4}

\int_{\frac{1}{7}}^{3} 14 x ln (7 x) d x

\frac{\partial}{\partial x} (\frac{5 y}{x ^{2} + 1})

\frac{d}{d x} (\frac{x + 1}{1 - x})

What is the integral from 2 to 7 of 4/(8x-11) ?
The integral from 2 to 7 of 4/(8x-11) is ln(3)