Calculus Examples

$\int_{1}^{x^{2} + 1} \frac{2 t + 2}{\sqrt{t + 1}} d t$

Step 1

Factor $2$ out of $2 t + 2$ .

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Step 1.1

Factor $2$ out of $2 t$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t) + 2}{\sqrt{t + 1}} d t]$

Step 1.2

Factor $2$ out of $2$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t) + 2 (1)}{\sqrt{t + 1}} d t]$

Step 1.3

Factor $2$ out of $2 (t) + 2 (1)$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1)}{\sqrt{t + 1}} d t]$

Step 2

Multiply $\frac{2 (t + 1)}{\sqrt{t + 1}}$ by $\frac{\sqrt{t + 1}}{\sqrt{t + 1}}$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1)}{\sqrt{t + 1}} \cdot \frac{\sqrt{t + 1}}{\sqrt{t + 1}} d t]$

Step 3

Simplify terms.

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Step 3.1

Combine and simplify the denominator.

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Step 3.1.1

Multiply $\frac{2 (t + 1)}{\sqrt{t + 1}}$ by $\frac{\sqrt{t + 1}}{\sqrt{t + 1}}$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{\sqrt{t + 1} \sqrt{t + 1}} d t]$

Step 3.1.2

Raise $\sqrt{t + 1}$ to the power of $1$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{\sqrt{t + 1}}^{1} \sqrt{t + 1}} d t]$

Step 3.1.3

Raise $\sqrt{t + 1}$ to the power of $1$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{\sqrt{t + 1}}^{1} {\sqrt{t + 1}}^{1}} d t]$

Step 3.1.4

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

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{\sqrt{t + 1}}^{1 + 1}} d t]$

Step 3.1.5

Add $1$ and $1$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{\sqrt{t + 1}}^{2}} d t]$

Step 3.1.6

Rewrite ${\sqrt{t + 1}}^{2}$ as $t + 1$ .

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Step 3.1.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{t + 1}$ as ${(t + 1)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{({(t + 1)}^{\frac{1}{2}})}^{2}} d t]$

Step 3.1.6.2

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

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{(t + 1)}^{\frac{1}{2} \cdot 2}} d t]$

Step 3.1.6.3

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

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{(t + 1)}^{\frac{2}{2}}} d t]$

Step 3.1.6.4

Cancel the common factor of $2$ .

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Step 3.1.6.4.1

Cancel the common factor.

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{(t + 1)}^{\frac{2}{2}}} d t]$

Step 3.1.6.4.2

Rewrite the expression.

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{{(t + 1)}^{1}} d t]$

Step 3.1.6.5

Simplify.

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{t + 1} d t]$

Step 3.2

Cancel the common factor of $t + 1$ .

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Step 3.2.1

Cancel the common factor.

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} \frac{2 (t + 1) \sqrt{t + 1}}{t + 1} d t]$

Step 3.2.2

Divide $2 \sqrt{t + 1}$ by $1$ .

$\frac{d}{d x} [\int_{1}^{x^{2} + 1} 2 \sqrt{t + 1} d t]$

Step 4

Take the derivative of $\int_{1}^{x^{2} + 1} 2 \sqrt{t + 1} d t$ with respect to $x$ using Fundamental Theorem of Calculus and the chain rule.

$\frac{d}{d x} [x^{2} + 1] (2 \sqrt{x^{2} + 1 + 1})$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$(\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]) (2 \sqrt{x^{2} + 1 + 1})$

Step 5.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$(2 x + \frac{d}{d x} [1]) (2 \sqrt{x^{2} + 1 + 1})$

Step 5.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$(2 x + 0) (2 \sqrt{x^{2} + 1 + 1})$

Step 5.4

Simplify the expression.

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Step 5.4.1

Add $2 x$ and $0$ .

$2 x (2 \sqrt{x^{2} + 1 + 1})$

Step 5.4.2

Add $1$ and $1$ .

$2 x (2 \sqrt{x^{2} + 2})$

Step 5.4.3

Multiply $2$ by $2$ .

$4 x \sqrt{x^{2} + 2}$