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

The integral from 0 to 1 of 1/(sqrt(1+2x)-1) is diverges

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integral from 0 to 1 of 1/(sqrt(1+2x)-1)

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

diverges

Solution steps

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

Multiply by the conjugate of

1 + 2 x - 1 : \frac{2 x}{1 + 2 x + 1}

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

Take the constant out:

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

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

Expand

\frac{1}{\frac{x}{1 + 2 x + 1}} : \frac{1 + 2 x}{x} + \frac{1}{x}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{1} \frac{1 + 2 x}{x} d x + \int_{0}^{1} \frac{1}{x} d x)

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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

\int_{- 1}^{2} 8 - x^{2} - x d x

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

\int_{- 2}^{1} 2 x^{2} d x

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