What is the integral from 1 to R of (ln(4x))/(x^2) ?

The integral from 1 to R of (ln(4x))/(x^2) is (-ln(4R)-1)/R+2ln(2)+1

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integral from 1 to R of (ln(4x))/(x^2)

\int_{1}^{R} \frac{ln ( 4 x )}{x ^{2}} d x

\frac{- ln ( 4 R ) - 1}{R} + 2 ln (2) + 1

Solution steps

\int_{1}^{R} \frac{ln ( 4 x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{R}

= [- \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{R}

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Compute the boundaries:

\frac{- ln ( 4 R ) - 1}{R} + 2 ln (2) + 1

= \frac{- ln ( 4 R ) - 1}{R} + 2 ln (2) + 1

\int \frac{e}{π} d x

\frac{\partial}{\partial x} (x^{2} e^{4 x y})

\frac{e ^{x}}{x ^{n}}

I n (x^{2})

\frac{d}{d x} (\frac{90}{6 + 9 e ^{- 0.025 x}})

What is the integral from 1 to R of (ln(4x))/(x^2) ?
The integral from 1 to R of (ln(4x))/(x^2) is (-ln(4R)-1)/R+2ln(2)+1