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

The integral from 1 to e of 1/(x(x^2+1)) is 1-1/2 (ln(e^2+1)-ln(2))

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

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

1 - \frac{1}{2} (ln (e^{2} + 1) - ln (2))

Decimal Notation

0.28310 \dots

Solution steps

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

Take the partial fraction of

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

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

\int_{1}^{e} \frac{1}{x} d x = 1

\int_{1}^{e} \frac{x}{x ^{2} + 1} d x = \frac{1}{2} (ln (e^{2} + 1) - ln (2))

= 1 - \frac{1}{2} (ln (e^{2} + 1) - ln (2))

\frac{d y}{d x} = y + x y^{3}

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

x^{^{''}} + 9 x = 0

n = 1 \sum \infty \frac{1}{1 + n ^{2}}

\frac{\partial}{\partial x} (2 x^{3} + y^{2})

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