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Found in: Page 334

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral : $$\int {\frac{{{x^4}}}{{x - 1}}dx}$$

First, divide $${x^4}$$ by $$x - 1$$, and then proceed further.

See the step by step solution

## Step 1:Given Data

$$\int {\frac{{{x^4}}}{{x - 1}}dx} = \int {{x^3} + {x^2} + x + 1 + \frac{1}{{x - 1}}dx}$$

## Step 2: Splitting the single integral into multiple integrals.

$$= \int {{x^3} dx + \int {{x^2}} dx + \int x dx + \int 1 dx + \int {\frac{1}{{x - 1}}} dx}$$

## Step 3: Solving each integral.

$$\begin{array}{l} = \frac{{{x^4}}}{4} + \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + \int {\frac{1}{{x - 1}}dx} \\\left( {Formula\,used:\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}}} } \right)\end{array}$$

## Step 4: Calculation of $$\int {\frac{1}{{x - 1}}dx}$$.

Let $$x - 1$$ be $$x$$

$$x - 1 = t$$

$$t = x - 1$$

$$\frac{{dt}}{{dx}} = 1 \Rightarrow dt = dx$$

Therefore, by substitution,

$$\begin{array}{l}\int {\frac{1}{{x - 1}}dx} = \int {\frac{1}{t}dt} \\ = log\left| t \right|\\ = log\left| {x - 1} \right|\end{array}$$

$$\int {\frac{{{x^4}}}{{x - 1}}} = \frac{{{x^4}}}{4} + \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + log\left| {x - 1} \right| + c$$

Hence, the value of the integral $$\int {\frac{{{x^4}}}{{x - 1}}} dx$$ is $$\frac{{{x^4}}}{4} + \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + log\left| {x - 1} \right| + c$$