What is the integral from 1 to e of 4/(x(10+9ln(x))) ?

The integral from 1 to e of 4/(x(10+9ln(x))) is 4/9 (ln(19)-ln(10))

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integral from 1 to e of 4/(x(10+9ln(x)))

\int_{1}^{e} \frac{4}{x ( 10 + 9 ln ( x ) )} d x

\frac{4}{9} (ln (19) - ln (10))

Decimal Notation

0.28526 \dots

Solution steps

\int_{1}^{e} \frac{4}{x ( 10 + 9 ln ( x ) )} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{e} \frac{1}{x ( 10 + 9 ln ( x ) )} d x

Apply u-substitution

: \int_{10}^{19} \frac{1}{9 u} d u

= 4 \cdot \int_{10}^{19} \frac{1}{9 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}{9} \cdot \int_{10}^{19} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{9} [ln ∣ u ∣]_{10}^{19}

Simplify

4 \cdot \frac{1}{9} [ln ∣ u ∣]_{10}^{19} : \frac{4}{9} [ln ∣ u ∣]_{10}^{19}

= \frac{4}{9} [ln ∣ u ∣]_{10}^{19}

Compute the boundaries:

ln (19) - ln (10)

= \frac{4}{9} (ln (19) - ln (10))

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What is the integral from 1 to e of 4/(x(10+9ln(x))) ?
The integral from 1 to e of 4/(x(10+9ln(x))) is 4/9 (ln(19)-ln(10))