Calculus Examples

$\int_{e}^{\infty} \frac{1}{x \ln^{2} (x)} d x$

Step 1

Write the integral as a limit as $t$ approaches $\infty$ .

$lim_{t \to \infty} \int_{e}^{t} \frac{1}{x \ln^{2} (x)} d x$

Step 2

Let $u = \ln (x)$ . Then $d u = \frac{1}{x} d x$ , so $x d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = \ln (x)$ . Find $\frac{d u}{d x}$ .

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Step 2.1.1

Differentiate $\ln (x)$ .

$\frac{d}{d x} [\ln (x)]$

Step 2.1.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{1}{x}$

Step 2.2

Substitute the lower limit in for $x$ in $u = \ln (x)$ .

$u_{lower} = \ln (e)$

Step 2.3

The natural logarithm of $e$ is $1$ .

$u_{lower} = 1$

Step 2.4

Substitute the upper limit in for $x$ in $u = \ln (x)$ .

$u_{upper} = \ln (t)$

Step 2.5

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = \ln (t)$

Step 2.6

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$lim_{t \to \infty} \int_{1}^{\ln (t)} \frac{1}{u^{2}} d u$

Step 3

Apply basic rules of exponents.

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Step 3.1

Move $u^{2}$ out of the denominator by raising it to the $- 1$ power.

$lim_{t \to \infty} \int_{1}^{\ln (t)} {(u^{2})}^{- 1} d u$

Step 3.2

Multiply the exponents in ${(u^{2})}^{- 1}$ .

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Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$lim_{t \to \infty} \int_{1}^{\ln (t)} u^{2 \cdot - 1} d u$

Step 3.2.2

Multiply $2$ by $- 1$ .

$lim_{t \to \infty} \int_{1}^{\ln (t)} u^{- 2} d u$

Step 4

By the Power Rule, the integral of $u^{- 2}$ with respect to $u$ is $- u^{- 1}$ .

$lim_{t \to \infty} - u^{- 1}]_{1}^{\ln (t)}$

Step 5

Substitute and simplify.

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Step 5.1

Evaluate $- u^{- 1}$ at $\ln (t)$ and at $1$ .

$lim_{t \to \infty} (- \ln^{- 1} (t)) + 1^{- 1}$

Step 5.2

One to any power is one.

$lim_{t \to \infty} - \ln^{- 1} (t) + 1$

Step 6

Evaluate the limit.

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Step 6.1

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $\infty$ .

$- lim_{t \to \infty} \ln^{- 1} (t) + lim_{t \to \infty} 1$

Step 6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- lim_{t \to \infty} \frac{1}{\ln (t)} + lim_{t \to \infty} 1$

Step 6.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{\ln (t)}$ approaches $0$ .

$- 0 + lim_{t \to \infty} 1$

Step 6.4

Evaluate the limit.

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Step 6.4.1

Evaluate the limit of $1$ which is constant as $t$ approaches $\infty$ .

$0 + 1$

Step 6.4.2

Add $0$ and $1$ .

$1$