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

The integral from 1 to 4 of ((ln(x)+4)^4)/x is 32/5 ln^5(2)+64ln^4(2)+256ln^3(2)+512ln^2(2)+512ln(2)

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integral from 1 to 4 of ((ln(x)+4)^4)/x

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

\frac{32}{5} ln^{5} (2) + 64 ln^{4} (2) + 256 ln^{3} (2) + 512 ln^{2} (2) + 512 ln (2)

Decimal Notation

701.93507 \dots

Solution steps

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

Apply Integration By Parts

: [(ln (x) + 4)^{4} - \int \frac{4 ( ln ( x ) + 4 ) ^{3} - ( ln ( x ) + 4 ) ^{4}}{x} d x]_{1}^{4}

= [(ln (x) + 4)^{4} - \int \frac{4 ( ln ( x ) + 4 ) ^{3} - ( ln ( x ) + 4 ) ^{4}}{x} d x]_{1}^{4}

\int \frac{4 ( ln ( x ) + 4 ) ^{3} - ( ln ( x ) + 4 ) ^{4}}{x} d x = (ln (x) + 4)^{4} - \frac{( ln ( x ) + 4 ) ^{5}}{5}

= [(ln (x) + 4)^{4} - ((ln (x) + 4)^{4} - \frac{( ln ( x ) + 4 ) ^{5}}{5})]_{1}^{4}

Simplify

[(ln (x) + 4)^{4} - ((ln (x) + 4)^{4} - \frac{( ln ( x ) + 4 ) ^{5}}{5})]_{1}^{4} : [\frac{1}{5} (ln (x) + 4)^{5}]_{1}^{4}

= [\frac{1}{5} (ln (x) + 4)^{5}]_{1}^{4}

Compute the boundaries:

\frac{1}{5} (2 ln (2) + 4)^{5} - \frac{1024}{5}

= \frac{1}{5} (2 ln (2) + 4)^{5} - \frac{1024}{5}

Simplify

= \frac{32}{5} ln^{5} (2) + 64 ln^{4} (2) + 256 ln^{3} (2) + 512 ln^{2} (2) + 512 ln (2)

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

What is the integral from 1 to 4 of ((ln(x)+4)^4)/x ?
The integral from 1 to 4 of ((ln(x)+4)^4)/x is 32/5 ln^5(2)+64ln^4(2)+256ln^3(2)+512ln^2(2)+512ln(2)