Question

Evaluate the integral :

\(\int {\frac{{{x^4}}}{{x - 1}}dx} \)

Short answer

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

Step-by-step solution

Step 1:Given Data

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

Splitting the single integral into multiple integrals.

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

Solving each integral.

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

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 \(\frac{{{x^4}}}{4} + \frac{{{x^3}}}{3} + \frac{{{x^2}}}{2} + x + log\left| {x - 1} \right| + c\)