What is the integral from 1 to 6 of (7ln(x))/(x^2) ?

The integral from 1 to 6 of (7ln(x))/(x^2) is 7((-ln(6)-1)/6+1)

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integral from 1 to 6 of (7ln(x))/(x^2)

\int_{1}^{6} \frac{7 ln ( x )}{x ^{2}} d x

7 (\frac{- ln ( 6 ) - 1}{6} + 1)

Decimal Notation

3.74294 \dots

Solution steps

\int_{1}^{6} \frac{7 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{1}^{6} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{6}

= 7 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{6}

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

= 7 [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{6}

Since the denominators are equal, combine the fractions:

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

= 7 [\frac{- ln ( x ) - 1}{x}]_{1}^{6}

Compute the boundaries:

\frac{- ln ( 6 ) - 1}{6} + 1

= 7 (\frac{- ln ( 6 ) - 1}{6} + 1)

x \to 8 lim (2 x - 3)

\int_{1}^{3} 4 r^{4} ln (r) d r

2^{x^{2} + x}

\int ∣ x - 2 ∣ d x

\frac{d y}{d x}, 3 x + 2 y = 5

What is the integral from 1 to 6 of (7ln(x))/(x^2) ?
The integral from 1 to 6 of (7ln(x))/(x^2) is 7((-ln(6)-1)/6+1)