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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 1

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 1

Answers (1)

Answer:

            \frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+C

Hint:

            To solve the given integration, we use partial fraction method

Given:

            \int \frac{2 x+1}{(x+1)(x-2)} d x

Explanation:

Let

I=\int \frac{2 x+1}{(x+1)(x-2)} d x

Expressing the function in terms of partial fraction as following

  \left.\frac{2 x+1}{(x+1)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x-2)} \quad \text { [Using } \frac{N(x)}{(a x+b)(c x+d)}=\frac{A}{a x+b}+\frac{B}{c x+d}\right]                   

\begin{aligned} &\frac{2 x+1}{(x+1)(x-2)}=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)} \\ &2 x+1=A(x-2)+B(x+1) \\ &2 x+1=(A+B) x+(-2 A+B) \end{aligned}

On comparing the coefficient of x and constant terms we get

\begin{gathered} 2=A+B \\ \end{gathered}                          (1)                         

\begin{aligned} -2 A+B=1 \end{aligned}                      (2)

Now, subtract equation (2) from (1)

\begin{gathered} A+B=2- \\ \frac{-2 A+B=1}{3 A=1} \\ A=\frac{1}{3} \end{gathered}

Putting the value of A in equation (1)

\begin{aligned} &\frac{1}{3}+B=2 \\ &B=2-\frac{1}{3} \\ &B=\frac{5}{3} \end{aligned}

Now

\begin{aligned} &\frac{2 x+1}{(x+1)(x-2)}=\frac{1}{3(x+1)}+\frac{5}{3(x-2)} \\ &I=\int\left(\frac{1}{3(x+1)}+\frac{5}{3(x-2)}\right) d x \\ &=\int \frac{1}{3(x+1)} d x+\int \frac{5}{3(x-2)} d x \end{aligned}

\left.=\frac{1}{3} \log |x+1| d x+\frac{5}{3} \log |x-2|+C \quad \text { [Using } \int \frac{1}{a x+b} d x=\frac{1}{a} \log |a x+b|+C\right]

Posted by

infoexpert27

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