What is the integral from 1 to 6 of 3/(x^2) ?

The integral from 1 to 6 of 3/(x^2) is 5/2

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integral from 1 to 6 of 3/(x^2)

\int_{1}^{6} \frac{3}{x ^{2}} d x

\frac{5}{2}

Decimal

2.5

Solution steps

\int_{1}^{6} \frac{3}{x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{6} \frac{1}{x ^{2}} d x

Apply the Power Rule

: [- \frac{1}{x}]_{1}^{6}

= 3 [- \frac{1}{x}]_{1}^{6}

Compute the boundaries:

\frac{5}{6}

= 3 \cdot \frac{5}{6}

Simplify

= \frac{5}{2}

y = - \frac{1}{5} x^{2} - \frac{6}{5} x + \frac{7}{5}, (- 5, 5)

\int e^{x} \cdot sin (2 x) d x

\frac{d}{d x} (3 5 x + 4 x^{\frac{1}{2}} - 3 x)

x^{2} \frac{d y}{d x} - 2 x y = 6 y^{4}, y (1) = \frac{1}{5}

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

What is the integral from 1 to 6 of 3/(x^2) ?
The integral from 1 to 6 of 3/(x^2) is 5/2