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  • Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 13 maths textbook solution

Answers (1)

Answer:

        \frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

        \int \frac{x}{3 x^{4}-18 x^{2}+11} d x

Solution:

Let\: \: I=\int \frac{x}{3 x^{4}-18 x^{2}+11} d x

              =\int \frac{x}{3 (x^{2})^{2}-18 x^{2}+11} d x

Put\: \: x^{2}=t\Rightarrow 2x\, dx=dt\Rightarrow x\, dx=\frac{dt}{2}

Then\: \: I=\int \frac{1}{3t^{2}-18 t+11} \, \frac{dt}{2}

                \begin{aligned} &=\frac{1}{2} \int \frac{1}{3\left(t^{2}-6 t+\frac{11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}-2.3 . t+(3)^{2}-(3)^{2}+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-9+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{27-11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\frac{16}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} d t \end{aligned}

Put\: \: t-3=u\Rightarrow dt=du

Then\: \: I =\frac{1}{6} \int \frac{1}{u^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} du

                \begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \cdot \frac{4}{\sqrt{3}}} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{\sqrt{3}}{6 \times 8} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \\ &=\frac{\sqrt{3}}{48} \log \left|\frac{t-3-\frac{4}{\sqrt{3}}}{t-3+\frac{4}{\sqrt{3}}}\right|+C \quad[\because u=t-3] \end{aligned}

                =\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C\: \: \: \: \: \: \: [\because t=x^{2}]

Posted by

Gurleen Kaur

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