Calculus Examples

$\int x {(x^{2} - 1)}^{48} d x$

Let $u = x^{2} - 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

$\int {(u)}^{48} \frac{1}{2} d u$

Combine fractions.

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Write ${(u)}^{48}$ as a fraction with denominator $1$ .

$\int \frac{{(u)}^{48}}{1} \frac{1}{2} d u$

Multiply $\frac{{(u)}^{48}}{1}$ and $\frac{1}{2}$ .

$\int \frac{{(u)}^{48}}{2} d u$

Since $\frac{1}{2}$ is constant with respect to $u$ , the integral of $\frac{{(u)}^{48}}{2}$ with respect to $u$ is $\frac{1}{2} \int {(u)}^{48} d u$ .

$\frac{1}{2} \int {(u)}^{48} d u$

By the Power Rule, the integral of ${(u)}^{48}$ with respect to $u$ is $\frac{1}{49} u^{49}$ .

$\frac{1}{2} (\frac{1}{49} u^{49} + C)$

Simplify the answer.

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Write $u^{49}$ as a fraction with denominator $1$ .

$\frac{1}{2} (\frac{1}{49} \frac{u^{49}}{1} + C)$

Multiply $\frac{1}{49}$ and $\frac{u^{49}}{1}$ .

$\frac{1}{2} (\frac{u^{49}}{49} + C)$

Simplify.

$\frac{1}{2} \frac{u^{49}}{49} + C$

Multiply $\frac{1}{2}$ and $\frac{u^{49}}{49}$ .

$\frac{u^{49}}{2 \cdot 49} + C$

Multiply $2$ by $49$ .

$\frac{u^{49}}{98} + C$

Replace all occurrences of $u$ with $x^{2} - 1$ .

$\frac{{(x^{2} - 1)}^{49}}{98} + C$

Simplify each term.

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Simplify the numerator.

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Rewrite $1$ as $1^{2}$ .

$\frac{{(x^{2} - 1^{2})}^{49}}{98} + C$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{{((x + 1) (x - 1))}^{49}}{98} + C$

Apply the product rule to $(x + 1) (x - 1)$ .

$\frac{{(x + 1)}^{49} {(x - 1)}^{49}}{98} + C$

Reorder terms.

$\frac{1}{98} {(x - 1)}^{49} {(x + 1)}^{49} + C$

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