What is the integral from 0 to 1 of 5/(sqrt(x)(1+x)) ?

The integral from 0 to 1 of 5/(sqrt(x)(1+x)) is (5pi)/2

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integral from 0 to 1 of 5/(sqrt(x)(1+x))

\int_{0}^{1} \frac{5}{x ( 1 + x )} d x

\frac{5 π}{2}

Decimal Notation

7.85398 \dots

Solution steps

\int_{0}^{1} \frac{5}{x ( 1 + x )} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{1}{x ( 1 + x )} d x

Apply u-substitution

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

= 5 \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

= 5 \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)

= 5 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 10 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 10 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{2}

\frac{d y}{d t} = k (80 - y) (50 - y)

\int \frac{1}{4 cos ( 2 x )} d x

x \to 4 lim (1 - 2 x)

\int_{0}^{1} t e^{2 t} d t

tan (x), at x = - \frac{π}{3}

What is the integral from 0 to 1 of 5/(sqrt(x)(1+x)) ?
The integral from 0 to 1 of 5/(sqrt(x)(1+x)) is (5pi)/2