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Popular Calculus >

integral from 1 to 2 of 1/x+y

Solution

\int_{1}^{2} \frac{1}{x} + y d x

Solution

ln (2) + y

Solution steps

\int_{1}^{2} \frac{1}{x} + y d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} \frac{1}{x} d x + \int_{1}^{2} y d x

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

\int_{1}^{2} y d x = y

= ln (2) + y

Popular Examples

integral from 8 to 10 of 0.5x^{-1.5}

integral from-2 to 2 of (1+2x)

integral from 1 to 2 of 1/((x-2)^{2/3)}

integral from 2 to 3 of (3t+2)

integral from 0 to 1 of (3/8)x^2

Frequently Asked Questions (FAQ)

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