What is the integral from 0 to 1 of 9/(sqrt(x)(1+x)) ?

The integral from 0 to 1 of 9/(sqrt(x)(1+x)) is (9pi)/2

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integral from 0 to 1 of 9/(sqrt(x)(1+x))

\int_{0}^{1} \frac{9}{x ( 1 + x )} d x

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{1} \frac{9}{x ( 1 + x )} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{1} \frac{1}{x ( 1 + x )} d x

Apply u-substitution

: \int_{0}^{1} \frac{2}{1 + u ^{2}} d u

= 9 \cdot \int_{0}^{1} \frac{2}{1 + u ^{2}} d u

Take the constant out:

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

= 9 \cdot 2 \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 9 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 18 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 18 \cdot \frac{π}{4}

Simplify

= \frac{9 π}{2}

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 5

\frac{d}{d x} (ln (\frac{1}{2 x}))

ln (3^{x})

y = 3 e^{x} cos (x)

\int (\frac{x}{7} - \frac{3}{4} x^{4}) d x

What is the integral from 0 to 1 of 9/(sqrt(x)(1+x)) ?
The integral from 0 to 1 of 9/(sqrt(x)(1+x)) is (9pi)/2