What is the integral from 1 to e^x of 4ln(x) ?

The integral from 1 to e^x of 4ln(x) is 4(e^xx-e^x+1)

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integral from 1 to e^x of 4ln(x)

\int_{1}^{e^{x}} 4 ln (x) d x

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

Solution steps

\int_{1}^{e^{x}} 4 ln (x) d x

Take the constant out:

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

= 4 \cdot \int_{1}^{e^{x}} ln (x) d x

Apply Integration By Parts

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

= 4 [x ln (x) - \int 1 d x]_{1}^{e^{x}}

\int 1 d x = x

= 4 [x ln (x) - x]_{1}^{e^{x}}

Compute the boundaries:

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

= 4 (e^{x} x - e^{x} + 1)

\int_{0}^{1} 2 x y d x

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

\int_{0}^{x} \frac{3}{x} d x

\int_{0}^{2} \frac{2 x}{x ^{2} + 4} d x

\int_{0}^{5} \frac{1}{5 - x} d x

What is the integral from 1 to e^x of 4ln(x) ?
The integral from 1 to e^x of 4ln(x) is 4(e^xx-e^x+1)