What is the integral from 1 to e^x of 9ln(t) ?

The integral from 1 to e^x of 9ln(t) is 9(xe^x-e^x+1)

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integral from 1 to e^x of 9ln(t)

\int_{1}^{e^{x}} 9 ln (t) d t

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

Solution steps

\int_{1}^{e^{x}} 9 ln (t) d t

Take the constant out:

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

= 9 \cdot \int_{1}^{e^{x}} ln (t) d t

Apply Integration By Parts

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

= 9 [t ln (t) - \int 1 d t]_{1}^{e^{x}}

\int 1 d t = t

= 9 [t ln (t) - t]_{1}^{e^{x}}

Compute the boundaries:

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

= 9 (x e^{x} - e^{x} + 1)

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What is the integral from 1 to e^x of 9ln(t) ?
The integral from 1 to e^x of 9ln(t) is 9(xe^x-e^x+1)