What is the integral from 3 to 4 of 1/(sqrt(x)-2x) ?

The integral from 3 to 4 of 1/(sqrt(x)-2x) is ln(-1+2sqrt(3))-ln(3)

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integral from 3 to 4 of 1/(sqrt(x)-2x)

\int_{3}^{4} \frac{1}{x - 2 x} d x

ln (- 1 + 23) - ln (3)

Decimal Notation

- 0.19678 \dots

Solution steps

\int_{3}^{4} \frac{1}{x - 2 x} d x

Apply u-substitution

: \int_{3}^{2} \frac{2}{1 - 2 u} d u

= \int_{3}^{2} \frac{2}{1 - 2 u} d u

Take the constant out:

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

= 2 \cdot \int_{3}^{2} \frac{1}{1 - 2 u} d u

Apply u-substitution

: \int_{1 - 23}^{- 3} - \frac{1}{2 v} d v

= 2 \cdot \int_{1 - 23}^{- 3} - \frac{1}{2 v} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 3}^{1 - 23} - \frac{1}{2 v} d v)

Take the constant out:

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

= 2 (- (- \frac{1}{2} \cdot \int_{- 3}^{1 - 23} \frac{1}{v} d v))

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- (- \frac{1}{2} [ln ∣ v ∣]_{- 3}^{1 - 23}))

Simplify

2 (- (- \frac{1}{2} [ln ∣ v ∣]_{- 3}^{1 - 23})) : [ln ∣ v ∣]_{- 3}^{1 - 23}

= [ln ∣ v ∣]_{- 3}^{1 - 23}

Compute the boundaries:

ln (- 1 + 23) - ln (3)

= ln (- 1 + 23) - ln (3)

\int_{0}^{4} \frac{x + 1}{x ^{2} + 2 x + 1} d x

\int_{0}^{0.5} e^{- x^{3}} d x

\int_{- 3}^{4} ∣ x + 1 ∣ d x

\int_{1}^{6} (x^{3} + x^{2} + 3) d x

\int_{2}^{3} \frac{7}{x} + x^{- 1} d x

What is the integral from 3 to 4 of 1/(sqrt(x)-2x) ?
The integral from 3 to 4 of 1/(sqrt(x)-2x) is ln(-1+2sqrt(3))-ln(3)