What is the integral from 1 to 6 of 5(ln(x))/(x^2) ?

The integral from 1 to 6 of 5(ln(x))/(x^2) is 5((-ln(6)-1)/6+1)

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integral from 1 to 6 of 5(ln(x))/(x^2)

\int_{1}^{6} 5 \frac{ln ( x )}{x ^{2}} d x

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

Decimal

2.67353 \dots

Solution steps

\int_{1}^{6} 5 \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

= 5 \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}

= 5 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{6}

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

= 5 [- \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}

= 5 [\frac{- ln ( x ) - 1}{x}]_{1}^{6}

Compute the boundaries:

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

= 5 (\frac{- ln ( 6 ) - 1}{6} + 1)

f (x) = cos (2 x)

f (x) = (x^{3} - 4)^{2} (2 x + 5)^{3}

\frac{d}{d x} (ln (5 x^{4} + 4 x))

\int \frac{6 x ^{2} + 5 x + 6}{( x ^{2} + 1 ) ^{2}} d x

y^{^{'}} = \frac{3 ^{2} y ^{2} + 1}{3}, y (0) = - 0.5

What is the integral from 1 to 6 of 5(ln(x))/(x^2) ?
The integral from 1 to 6 of 5(ln(x))/(x^2) is 5((-ln(6)-1)/6+1)