What is the integral from 1 to e^2 of ln(x) ?

The integral from 1 to e^2 of ln(x) is e^2+1

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integral from 1 to e^2 of ln(x)

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

e^{2} + 1

Decimal Notation

8.38905 \dots

Solution steps

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

Apply Integration By Parts

: [x ln (x) - \int 1 d x]_{1}^{e^{2}}

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

\int 1 d x = x

= [x ln (x) - x]_{1}^{e^{2}}

Compute the boundaries:

e^{2} + 1

= e^{2} + 1

y^{^{''}} + 14 y^{^{'}} + 58 y = 0

\frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

\int 2 (x - 2) \cdot ln (x - 2) d x

\int \frac{1}{4 - 6 x} d x

d x + d y + x d y = y d x

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