What is the integral from 1 to 2 of 4/(2x+3) ?

The integral from 1 to 2 of 4/(2x+3) is 2(ln(7)-ln(5))

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integral from 1 to 2 of 4/(2x+3)

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

2 (ln (7) - ln (5))

Decimal

0.67294 \dots

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{1}^{2} \frac{1}{2 x + 3} d x

Apply u-substitution

: \int_{5}^{7} \frac{1}{2 u} d u

= 4 \cdot \int_{5}^{7} \frac{1}{2 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}{2} \cdot \int_{5}^{7} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{7}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{7} : 2 [ln ∣ u ∣]_{5}^{7}

= 2 [ln ∣ u ∣]_{5}^{7}

Compute the boundaries:

ln (7) - ln (5)

= 2 (ln (7) - ln (5))

2 (t - 2)

\int \frac{1}{16 x ^{2} - 24 x - 27} d x

\frac{x}{5}

y = 7 cos (x)

\int t^{2} \cdot ln^{2} (t) d t

What is the integral from 1 to 2 of 4/(2x+3) ?
The integral from 1 to 2 of 4/(2x+3) is 2(ln(7)-ln(5))