What is the integral from 1 to x of ln(y) ?

The integral from 1 to x of ln(y) is xln(x)-x+1

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integral from 1 to x of ln(y)

\int_{1}^{x} ln (y) d y

x ln (x) - x + 1

Solution steps

\int_{1}^{x} ln (y) d y

Apply Integration By Parts

: [y ln (y) - \int 1 d y]_{1}^{x}

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

\int 1 d y = y

= [y ln (y) - y]_{1}^{x}

Compute the boundaries:

x ln (x) - x + 1

= x ln (x) - x + 1

\int_{0}^{\frac{1}{2}} (cos (π x) - 2^{x}) d x

\int_{0}^{1} (e^{t} - e + 1)^{2} d t

\int_{0}^{1} 1

\int_{1}^{2} (2 - 6 y^{2}) d y

\int_{2}^{5} 2 π y (5 - y) d y

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