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

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

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

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

7 (\frac{- ln ( 2 ) - 1}{2} + 1)

Decimal

1.07398 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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}^{2}

Compute the boundaries:

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

= 7 (\frac{- ln ( 2 ) - 1}{2} + 1)

\int 5 x \cdot 3 1 - x^{2} d x

\frac{\partial}{\partial x} (\frac{1}{1 - b x})

\frac{d}{d x} ((2 x - 7)^{4} (x^{2} + x + 1)^{5})

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

x \to \infty lim (- \frac{sin ^{2} ( x ) + 2}{2 x - 5})

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