What is the integral from 0 to 1 of 7/(sqrt(x)(x+1)) ?

The integral from 0 to 1 of 7/(sqrt(x)(x+1)) is (7pi)/2

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integral from 0 to 1 of 7/(sqrt(x)(x+1))

\int_{0}^{1} \frac{7}{x ( x + 1 )} d x

\frac{7 π}{2}

Decimal Notation

10.99557 \dots

Solution steps

\int_{0}^{1} \frac{7}{x ( x + 1 )} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} \frac{1}{x ( x + 1 )} d x

Apply u-substitution

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

= 7 \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

= 7 \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)

= 7 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 14 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 14 \cdot \frac{π}{4}

Simplify

= \frac{7 π}{2}

\int_{3}^{5} \frac{1}{x - 4} d x

y = (x^{2} - 4 x)^{5}

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

(x^{x^{x^{x}}})^{^{'}}

(7 x + 6 y) d x + (6 x - 8 y 3) d y = 0

What is the integral from 0 to 1 of 7/(sqrt(x)(x+1)) ?
The integral from 0 to 1 of 7/(sqrt(x)(x+1)) is (7pi)/2