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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 123 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 123 Maths Textbook Solution.

Answers (1)

Answer:=\frac{1}{3} \log |x+1|-\frac{1}{6} \log \left|x^{2}+x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1} \frac{2 x+1}{\sqrt{3}}+C

Given: \int \frac{x}{x^{3}-1} d x

Hint: using partial function

Explanation let I=\int \frac{x}{x^{3}-1} d x

\frac{x}{x^{3}-1}=\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+c}{x^{2}+x+1}

Multiplying by  (x-1)\left(x^{2}+x+1\right)

x=A\left(x^{2}+x+1\right)+(B x+C)(x-1)

Putting x=1

\begin{aligned} &1=A(1+1+1)+(B(1)+C)(1-1) \\ &1=3 A+0 \Rightarrow A=\frac{1}{3} \end{aligned}

Putting x=-1

\begin{aligned} &-1=A+(-2)(-B+C)\\ &-1=\frac{1}{3}+2 B-2 C\\ &2(B-C)=-1 \frac{-1}{3} \Rightarrow B-C=\frac{-2}{3}\\ &B-C=>\frac{-2}{3} \end{aligned}

Putting x=0

\begin{aligned} &0=A(0+0+1)+(0+C)(0-1) \\ &0=A(1)+C(-1) \\ &0=\frac{1}{3}-C=>C=\frac{1}{3} \end{aligned}

Put in (1)

\begin{aligned} &B-\frac{1}{3}=\frac{-2}{3}=>B=\frac{1}{3} \frac{-2}{3}=>B=\frac{-1}{3} \\ &\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3(x-1)}+\frac{\frac{-1}{3} x+\frac{1}{3}}{x^{2}+x+1} \\ &\int \frac{x d x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x-1}{x^{2}+x+1} d x \\ &=\frac{1}{3} \log |x-1|-\frac{1}{3} I_{1} \quad \text { (2) } \end{aligned}

Where I_{1}=\int \frac{x-1}{x^{2}+x+1} d x

\begin{aligned} &=\frac{1}{2} \int \frac{2(x-1)+3-3}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+1} d x-\frac{3}{2} \int \frac{1}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x \quad\left[\begin{array}{l} x^{2}+x+1=t \\ (2 x+1) d x=d t \end{array}\right] \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &=\frac{1}{2} \log |t|-\frac{3}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \\ &=\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}

By (2), we get

\begin{aligned} &=\frac{1}{3} \log |x-1|-\frac{1}{3}\left[\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]+C \\ &=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^{2}+x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}

 

 

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