Calculus Examples

$\int_{0}^{1} \frac{\ln (x)}{\sqrt{x}} d x$

Step 1

Rewrite $\frac{\ln (x)}{\sqrt{x}}$ as a product.

$\int_{0}^{1} \ln (x) \frac{1}{\sqrt{x}} d x$

Step 2

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x)$ and $d v = \frac{1}{\sqrt{x}}$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} 2 x^{\frac{1}{2}} \frac{1}{x} d x$

Step 3

Simplify.

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Step 3.1

Combine $2$ and $\frac{1}{x}$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} x^{\frac{1}{2}} \frac{2}{x} d x$

Step 3.2

Combine $x^{\frac{1}{2}}$ and $\frac{2}{x}$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{x^{\frac{1}{2}} \cdot 2}{x} d x$

Step 3.3

Move $2$ to the left of $x^{\frac{1}{2}}$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2 \cdot x^{\frac{1}{2}}}{x} d x$

Step 3.4

Move $x^{\frac{1}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x \cdot x^{- \frac{1}{2}}} d x$

Step 3.5

Multiply $x$ by $x^{- \frac{1}{2}}$ by adding the exponents.

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Step 3.5.1

Multiply $x$ by $x^{- \frac{1}{2}}$ .

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Step 3.5.1.1

Raise $x$ to the power of $1$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x^{1} x^{- \frac{1}{2}}} d x$

Step 3.5.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x^{1 - \frac{1}{2}}} d x$

Step 3.5.2

Write $1$ as a fraction with a common denominator.

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x^{\frac{2}{2} - \frac{1}{2}}} d x$

Step 3.5.3

Combine the numerators over the common denominator.

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x^{\frac{2 - 1}{2}}} d x$

Step 3.5.4

Subtract $1$ from $2$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - \int_{0}^{1} \frac{2}{x^{\frac{1}{2}}} d x$

Step 4

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - (2 \int_{0}^{1} \frac{1}{x^{\frac{1}{2}}} d x)$

Step 5

Simplify the expression.

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Step 5.1

Multiply $2$ by $- 1$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 \int_{0}^{1} \frac{1}{x^{\frac{1}{2}}} d x$

Step 5.2

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

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 \int_{0}^{1} {(x^{\frac{1}{2}})}^{- 1} d x$

Step 5.3

Multiply the exponents in ${(x^{\frac{1}{2}})}^{- 1}$ .

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Step 5.3.1

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

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 \int_{0}^{1} x^{\frac{1}{2} \cdot - 1} d x$

Step 5.3.2

Combine $\frac{1}{2}$ and $- 1$ .

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 \int_{0}^{1} x^{\frac{- 1}{2}} d x$

Step 5.3.3

Move the negative in front of the fraction.

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 \int_{0}^{1} x^{- \frac{1}{2}} d x$

Step 6

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

$\ln (x) (2 x^{\frac{1}{2}})]_{0}^{1} - 2 (2 x^{\frac{1}{2}}]_{0}^{1})$

Step 7

Evaluate $\ln (x) (2 x^{\frac{1}{2}})$ at $1$ and at $0$ .

$(\ln (1) (2 \cdot 1^{\frac{1}{2}})) - \ln (0) (2 x^{\frac{1}{2}}) - 2 (2 x^{\frac{1}{2}}]_{0}^{1})$

Step 8

The natural logarithm of zero is undefined.

Undefined