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

The integral from b to 1 of 2ln(x) is 2(-1-bln(b)+b)

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integral from b to 1 of 2ln(x)

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

2 (- 1 - b ln (b) + b)

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{b}^{1} ln (x) d x

Apply Integration By Parts

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

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

\int 1 d x = x

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

Compute the boundaries:

- 1 - b ln (b) + b

= 2 (- 1 - b ln (b) + b)

\int (x^{2} + 2)^{\frac{3}{2}} d x

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{5}}})

\int \frac{x ^{3}}{( 4 - x ^{2} ) ^{3}} d x

y^{^{'}} = (x + y - 1)^{2} - 1

\frac{\partial}{\partial x} (\frac{t}{x - 1})

What is the integral from b to 1 of 2ln(x) ?
The integral from b to 1 of 2ln(x) is 2(-1-bln(b)+b)