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

The integral from-1 to u of 1/(x+x^2) is diverges

Popular >

integral from-1 to u of 1/(x+x^2)

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

diverges

Solution steps

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

Take the partial fraction of

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

= \int_{- 1}^{u} \frac{1}{x} - \frac{1}{x + 1} 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

= \int_{- 1}^{u} \frac{1}{x} d x - \int_{- 1}^{u} \frac{1}{x + 1} d x

\int_{- 1}^{u} \frac{1}{x} d x = ln ∣ u ∣

\int_{- 1}^{u} \frac{1}{x + 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{2}^{5} 5 d x

\int_{0}^{3} x^{3} - 4 x d x

\int_{0}^{l n (3)} 3 e^{3 x} d x

\int_{0}^{1} y (e^{y} - 1) d y

\int_{0}^{4} - x^{2} + 2 x d x

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