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

The integral from 1 to n of ln(x) is nln(n)-n+1

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integral from 1 to n of ln(x)

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

n ln (n) - n + 1

Solution steps

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

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

n ln (n) - n + 1

= n ln (n) - n + 1

\frac{d}{d x} (\frac{9 + 4 x - 6 x}{x})

\frac{\partial}{\partial x} (ln (x \cdot cos (y)))

\int \frac{1}{( x ^{2} - 441 ) ^{\frac{3}{2}}} d x

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

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{- 2 t}

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