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

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

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

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

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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}{d x} (3 x arctan (x^{2}) + e^{s i n (x)})

\int 3 x x + 5 d x

\frac{d}{d x} (x ln (x^{3}))

\int_{1}^{\infty} x e^{- 14 x} d x

13 (t + 1) \frac{d y}{d t} - 9 y = 36 t

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