Calculus Examples

$\int \frac{t}{t + 1} d t$

Step 1

Divide $t$ by $t + 1$ .

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$t$

$1$

$t$

$0$

Step 1.2

Divide the highest order term in the dividend $t$ by the highest order term in divisor $t$ .

					$1$
	$t$	+	$1$		$t$	+	$0$

Step 1.3

Multiply the new quotient term by the divisor.

				$1$
$t$	+	$1$		$t$	+	$0$
			+	$t$	+	$1$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $t + 1$

				$1$
$t$	+	$1$		$t$	+	$0$
			-	$t$	-	$1$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$t$	+	$1$		$t$	+	$0$
			-	$t$	-	$1$
					-	$1$

Step 1.6

The final answer is the quotient plus the remainder over the divisor.

$\int 1 - \frac{1}{t + 1} d t$

Step 2

Split the single integral into multiple integrals.

$\int d t + \int - \frac{1}{t + 1} d t$

Step 3

Apply the constant rule.

$t + C + \int - \frac{1}{t + 1} d t$

Step 4

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$t + C - \int \frac{1}{t + 1} d t$

Step 5

Let $u = t + 1$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = t + 1$ . Find $\frac{d u}{d t}$ .

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Step 5.1.1

Differentiate $t + 1$ .

$\frac{d}{d t} [t + 1]$

Step 5.1.2

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

$\frac{d}{d t} [t] + \frac{d}{d t} [1]$

Step 5.1.3

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

$1 + \frac{d}{d t} [1]$

Step 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Rewrite the problem using $u$ and $d u$ .

$t + C - \int \frac{1}{u} d u$

Step 6

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$t + C - (\ln (| u |) + C)$

Step 7

Simplify.

$t - \ln (| u |) + C$

Step 8

Replace all occurrences of $u$ with $t + 1$ .

$t - \ln (| t + 1 |) + C$