What is the integral from 1 to e^x of 7ln(x) ?

The integral from 1 to e^x of 7ln(x) is 7(e^xx-e^x+1)

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integral from 1 to e^x of 7ln(x)

\int_{1}^{e^{x}} 7 ln (x) d x

7 (e^{x} x - e^{x} + 1)

Solution steps

\int_{1}^{e^{x}} 7 ln (x) d x

Take the constant out:

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

= 7 \cdot \int_{1}^{e^{x}} ln (x) d x

Apply Integration By Parts

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

= 7 [x ln (x) - \int 1 d x]_{1}^{e^{x}}

\int 1 d x = x

= 7 [x ln (x) - x]_{1}^{e^{x}}

Compute the boundaries:

e^{x} x - e^{x} + 1

= 7 (e^{x} x - e^{x} + 1)

\int_{3}^{4} e^{x} d x

\int_{1}^{3} 4^{3 x} d x

\int_{3}^{5} sin (I n x) d x

\int_{0}^{4} x + 1 d x

\int_{0}^{2} (6 x - 3) (4 x^{2} + 7) d x

What is the integral from 1 to e^x of 7ln(x) ?
The integral from 1 to e^x of 7ln(x) is 7(e^xx-e^x+1)