What is the integral from 1 to pi of ln(t) ?

The integral from 1 to pi of ln(t) is piln(pi)-pi+1

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integral from 1 to pi of ln(t)

\int_{1}^{π} ln (t) d t

π ln (π) - π + 1

Decimal Notation

1.45468 \dots

Solution steps

\int_{1}^{π} ln (t) d t

Apply Integration By Parts

: [t ln (t) - \int 1 d t]_{1}^{π}

= [t ln (t) - \int 1 d t]_{1}^{π}

\int 1 d t = t

= [t ln (t) - t]_{1}^{π}

Compute the boundaries:

π ln (π) - π + 1

= π ln (π) - π + 1

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

\int_{5}^{10} \frac{1}{t + 1} d t

\int_{1}^{2} 1 + 64 x^{2} d x

What is the integral from 1 to pi of ln(t) ?
The integral from 1 to pi of ln(t) is piln(pi)-pi+1