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

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

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

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

\frac{7 π}{2}

Decimal Notation

10.99557 \dots

Solution steps

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

Apply u-substitution

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

= 7 \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

= 7 \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)

= 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}

\frac{\partial}{\partial y} (x)

x \to 0 lim ((x^{2}) cos (\frac{8}{x}))

\frac{\partial}{\partial x} (- 10 x y sin (x y^{2}))

f (x) = (2 x - x^{3}) (3 x + 7)

\frac{d}{d x} (5 x^{2} + 1)

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