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Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 36 Maths Textbook Solution.

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            \frac{1}{2}log\; \; 6

Hint: Use indefinite formula and put limits to solve this integral


\int_{1}^{2}\frac{x+3}{x\left ( x+2 \right )}dx


\begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\int_{1}^{2} \frac{x}{x+2} d x+\int_{1}^{2} \frac{3}{x(x+2)} d x \\ &=\int_{1}^{2} \frac{1}{x+2} d x+3 \int_{1}^{2} \frac{1}{x(x+2)} d x \end{aligned}                ..............(1)



Putting x+2=u
\Rightarrow dx=du 

when x=1 then u=1+2=3and when x=2 then u=2+2=4


\begin{aligned} \int_{1}^{2} \frac{1}{x+2} d x &=\int_{3}^{4} \frac{1}{u} d u=[\log u]_{3}^{4} \quad\left[\because \int \frac{1}{x} d x=\log x\right] \\ &=[\log 4-\log 3] \end{aligned}                ...............(2)


\int_{1}^{2}\frac{1}{x\left ( x+2 \right )}dx

To solve this integral, first we need to find its partial fraction then integrate it and put the given limits. So

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

Equating coefficient of x from both sides, then

0=A+B\Rightarrow B=-A

Again, Equating coefficient of constant term from both sides then

1=2A\Rightarrow A=\frac{1}{2}

\Rightarrow B=-\frac{1}{2}

\therefore A=\frac{1}{2}&B=-\frac{1}{2}


\begin{aligned} &\frac{1}{x(x+2)}=\frac{1}{2 x}-\frac{1}{2(x+2)} \\ &\therefore \int_{1}^{2} \frac{1}{x(x+2)} d x=\int_{1}^{2}\left(\frac{1}{2 x}-\frac{1}{2(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{(x+2)} d x \end{aligned}

Put x+2=u

\Rightarrow dx=du in the 2nd integral then

\begin{aligned} &\int_{1}^{2} \frac{1}{x(x+2)} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{u} d x \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log u]_{1}^{2} \\ &{\left[\because \int \frac{1}{x} d x=\log x\right]} \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ &{[\because u=x+2]} \end{aligned}

\begin{aligned} &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}[\log (2+2)-\log (1+2)] \\ &=\frac{1}{2}[\log 2-0]-\frac{1}{2}[\log (4)-\log (3)] \\ &=\frac{1}{2}[\log 2]-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \end{aligned}                                    ..............(3)

Putting the value of integrals from eq(2) and (3) in (1), we get

\begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\log 4-\log 3+3\left[\frac{1}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3\right] \\ &=\log 4-\log 3+\frac{3}{2} \log 2-\frac{3}{2} \log 4+\frac{3}{2} \log 3 \\ &=\left(1-\frac{3}{2}\right) \log 4-\left(1-\frac{3}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\left(\frac{2-3}{2}\right) \log 4-\left(\frac{2-3}{2}\right) \log 3+\frac{3}{2} \log 2 \end{aligned}

\begin{aligned} &=-\frac{1}{2} \log 4-\left(-\frac{1}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\frac{1}{2}\left[-\log 2^{2}+\log 3+3 \log 2\right] \\ &=\frac{1}{2}[-2 \log 2+\log 3+3 \log 2] \\ &=\frac{1}{2}[\log 2+\log 3] \end{aligned}

$$ \begin{aligned} &=\frac{1}{2} \log (2 \times 3) \\ &{[\because \log (m n)=\log m+\log n]} \\ &=\frac{1}{2} \log 6 \end{aligned}


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