What is the integral from 0 to 2 of 1/((3x+4)) ?

The integral from 0 to 2 of 1/((3x+4)) is 1/3 ln(5/2)

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integral from 0 to 2 of 1/((3x+4))

\int_{0}^{2} \frac{1}{( 3 x + 4 )} d x

\frac{1}{3} ln (\frac{5}{2})

Decimal Notation

0.30543 \dots

Solution steps

\int_{0}^{2} \frac{1}{3 x + 4} d x

Apply u-substitution

: \int_{4}^{10} \frac{1}{3 u} d u

= \int_{4}^{10} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{4}^{10}

Compute the boundaries:

ln (10) - 2 ln (2)

= \frac{1}{3} (ln (10) - 2 ln (2))

Simplify

= \frac{1}{3} ln (\frac{5}{2})

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

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\int_{0}^{12} \frac{1}{1 + x ^{2}} d x

\int_{3}^{\infty} \frac{3}{x ^{6}} d x

What is the integral from 0 to 2 of 1/((3x+4)) ?
The integral from 0 to 2 of 1/((3x+4)) is 1/3 ln(5/2)