# How to Find the Definite Integral of a Rational Function

## Definite integral math problem

How to find the definite Integral of the rational function “root x divided by 1 minus x” from 0 to 9

\begin{aligned} \int_{0}^{9}\dfrac{\sqrt{x}}{1-x} \ dx =?\\ \end{aligned}

## Solution to the definite integral math problem

Let

\begin{aligned} I=\int_{0}^{9}\dfrac{\sqrt{x}}{1-x} \ dx\\ \end{aligned}

The integral involves √x term so Let substitute **u = √x** then

du = dx/(2√x)

⇒ du = dx/(2u)

Thus, dx = 2udu

Limits of the integrals are 0 and 9

When x = 0 ⇒ **u **= √x = √0 =** 0**

When x = 9 ⇒ **u **= √x = √9 =** 3**

so new limits are 0 and 3

\begin{aligned} I&=\int_{0}^{9}\dfrac{\sqrt{x}}{1-x} \ dx \\ \\ &=\int_{0}^{3}\dfrac{u}{1-u^2} \times 2u du \\ \\ &=2\int_{0}^{3}\dfrac{u^2}{1-u^2} \ du \\ \\ \end{aligned}

Write u² = u² + 1 – 1, then

\begin{aligned} I &= 2\int_{0}^{3}\dfrac{u^2-1+1}{1-u^2} \ du \\ \\ &=2\int_{0}^{3}\dfrac{u^2-1}{1-u^2} \ du + 2\int_{0}^{3}\dfrac{ 1}{1-u^2} \ du\\ \\ &=2\int_{0}^{3}-1 \ du + 2\int_{0}^{3}\dfrac{ 1}{1-u^2} \ du\\ \\ \Rightarrow I&=I_1+I_2 \end{aligned}

## Solving integral I_{1}

\begin{aligned} I_1&= 2\int_{0}^{3}-1 \ du \\ \\ &=-2\int_{0}^{3}1 \ du \\ \\ &=-2\bigg[u \bigg]_{0}^{3} \ du \\ \\ &=-2(3 - 0) \\ \\ \Rightarrow I_1&=-6 \\ \end{aligned}

## Solving integral I_{2} using partial fraction

I_2=2\int_{0}^{3}\dfrac{ 1}{1-u^2} \ du

Factorise the denomenator

I_2=2\int_{0}^{3}\dfrac{ 1}{(1+u)(1-u)} \ du

### Apply partial fraction

let

\begin{aligned} \dfrac{ 1}{(1+u)(1-u)} &= \dfrac{A}{1+u}+\dfrac{B}{1-u} \\ \\ &= \dfrac{A(1-u)+B(1+u)}{(1+u)(1-u)} \\ \\ \Rightarrow 1&=A(1-u)+B(1+u) \\ \\ &=A-Au+B+Bu \\ \\ \Rightarrow 1&=(B-A)u+(A+B) \\ \\ \Rightarrow B-A&=0 \ \ \& \ A+B = 1\\ \\ \Rightarrow A=&B=\frac{1}{2} \\ \end{aligned}

Now substitute A and B

\begin{aligned} I_2&=2\int_{0}^{3}\dfrac{ 1}{(1+u)(1-u)} \ du \\ \\ &=2\int_{0}^{3}\bigg(\dfrac{A}{1+u} + \dfrac{B}{1-u}\bigg) \ du \\ \\ &=2\int_{0}^{3}\dfrac{A}{1+u} +2\int_{0}^{3}\dfrac{B}{1-u} \ du \\ \\ &=2\int_{0}^{3}\dfrac{\dfrac{1}{2}}{1+u} +2\int_{0}^{3}\dfrac{\dfrac{1}{2}}{1-u} \ du \\ \\ &=\int_{0}^{3}\dfrac{1}{1+u} +\int_{0}^{3}\dfrac{1}{1-u} \ du \\ \\ &=\bigg[\ln|1+u|\bigg]_{0}^{3} +\bigg[-\ln|1-u|\bigg]_{0}^{3} \ du \\ \\ &=\ln4+ \ln1 -\ln2 + \ln1 \\ \\ \Rightarrow I_2&=\ln4-\ln2 \\ \\ \end{aligned}

So

\begin{aligned} I&=I_1+I_2\\ \\ \Rightarrow I&= \ln4-\ln2 -6 \\ \end{aligned}

Now we can write

\begin{aligned} \int_{0}^{9}\dfrac{\sqrt{x}}{1-x} \ dx &= \ln4-\ln2 -6\\ \end{aligned}