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

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

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

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

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

Solution steps

\int_{1}^{6} 7 \cdot \frac{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)

\int sin (2 x) e^{c o s^{2} (x)} d x

x \to 3 lim (\frac{6}{x - 3})

2 x - 3

x \to - 1 - lim (\frac{1}{x + 1})

\frac{d y}{d x} + y = x y^{8}

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