(D) : Given, Let, \(g\left(\frac{t+1}{2 t+1}\right)=t+1\) \(\frac{t+1}{2 t+1}=x \Rightarrow t+1=x(2 t+1)\) \(t-2 t x=x-1 \Rightarrow t(1-2 x)=x-1\) \(t=\frac{x-1}{1-2 x}\) \(\text { So, } f(x)=\frac{x-1}{1-2 x}+1=\frac{x-1+1-2 x}{1-2 x}=\frac{x}{2 x-1}\) \(\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x\) \(=\frac{1}{2}\left[\int \frac{2 x-1}{2 x-1} d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[\int d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[x+\frac{1}{2} \log |2 x-1|\right]+C\) \(=\frac{x}{2}+\frac{1}{4} \log |2 x-1|+C\)
AP EAMCET-2018-24.04.2018
Integral Calculus
86147
\(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x=\)
1 \(\frac{x^{5}}{4}+x^{3}+6 x^{2}+c\)
2 \(\frac{x^{5}}{5}+\frac{x^{4}}{4}+6 x+c\)
3 \(\frac{x^{5}}{5}+x^{3}+6 x+c\)
4 \(\frac{x^{5}}{5}-\frac{x^{3}}{2}+6 x^{2}+c\)
Explanation:
(C) : Given, \(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x\) \(x^{8}-9 x^{2}+18\) is divided by \(x^{4}-3 x^{2}+3\), we get quotient \(\mathrm{x}^{4}+3 \mathrm{x}^{2}+6\) and 0 remainder. Now, \(\quad \int\left(x^{4}+3 x^{2}+6\right) d x\) \(=\frac{x^{5}}{5}+\frac{3 x^{3}}{3}+6 x+C=\frac{x^{5}}{5}+x^{3}+6 x+C\)
(D) : Given, Let, \(g\left(\frac{t+1}{2 t+1}\right)=t+1\) \(\frac{t+1}{2 t+1}=x \Rightarrow t+1=x(2 t+1)\) \(t-2 t x=x-1 \Rightarrow t(1-2 x)=x-1\) \(t=\frac{x-1}{1-2 x}\) \(\text { So, } f(x)=\frac{x-1}{1-2 x}+1=\frac{x-1+1-2 x}{1-2 x}=\frac{x}{2 x-1}\) \(\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x\) \(=\frac{1}{2}\left[\int \frac{2 x-1}{2 x-1} d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[\int d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[x+\frac{1}{2} \log |2 x-1|\right]+C\) \(=\frac{x}{2}+\frac{1}{4} \log |2 x-1|+C\)
AP EAMCET-2018-24.04.2018
Integral Calculus
86147
\(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x=\)
1 \(\frac{x^{5}}{4}+x^{3}+6 x^{2}+c\)
2 \(\frac{x^{5}}{5}+\frac{x^{4}}{4}+6 x+c\)
3 \(\frac{x^{5}}{5}+x^{3}+6 x+c\)
4 \(\frac{x^{5}}{5}-\frac{x^{3}}{2}+6 x^{2}+c\)
Explanation:
(C) : Given, \(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x\) \(x^{8}-9 x^{2}+18\) is divided by \(x^{4}-3 x^{2}+3\), we get quotient \(\mathrm{x}^{4}+3 \mathrm{x}^{2}+6\) and 0 remainder. Now, \(\quad \int\left(x^{4}+3 x^{2}+6\right) d x\) \(=\frac{x^{5}}{5}+\frac{3 x^{3}}{3}+6 x+C=\frac{x^{5}}{5}+x^{3}+6 x+C\)
(D) : Given, Let, \(g\left(\frac{t+1}{2 t+1}\right)=t+1\) \(\frac{t+1}{2 t+1}=x \Rightarrow t+1=x(2 t+1)\) \(t-2 t x=x-1 \Rightarrow t(1-2 x)=x-1\) \(t=\frac{x-1}{1-2 x}\) \(\text { So, } f(x)=\frac{x-1}{1-2 x}+1=\frac{x-1+1-2 x}{1-2 x}=\frac{x}{2 x-1}\) \(\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x\) \(=\frac{1}{2}\left[\int \frac{2 x-1}{2 x-1} d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[\int d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[x+\frac{1}{2} \log |2 x-1|\right]+C\) \(=\frac{x}{2}+\frac{1}{4} \log |2 x-1|+C\)
AP EAMCET-2018-24.04.2018
Integral Calculus
86147
\(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x=\)
1 \(\frac{x^{5}}{4}+x^{3}+6 x^{2}+c\)
2 \(\frac{x^{5}}{5}+\frac{x^{4}}{4}+6 x+c\)
3 \(\frac{x^{5}}{5}+x^{3}+6 x+c\)
4 \(\frac{x^{5}}{5}-\frac{x^{3}}{2}+6 x^{2}+c\)
Explanation:
(C) : Given, \(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x\) \(x^{8}-9 x^{2}+18\) is divided by \(x^{4}-3 x^{2}+3\), we get quotient \(\mathrm{x}^{4}+3 \mathrm{x}^{2}+6\) and 0 remainder. Now, \(\quad \int\left(x^{4}+3 x^{2}+6\right) d x\) \(=\frac{x^{5}}{5}+\frac{3 x^{3}}{3}+6 x+C=\frac{x^{5}}{5}+x^{3}+6 x+C\)
(D) : Given, Let, \(g\left(\frac{t+1}{2 t+1}\right)=t+1\) \(\frac{t+1}{2 t+1}=x \Rightarrow t+1=x(2 t+1)\) \(t-2 t x=x-1 \Rightarrow t(1-2 x)=x-1\) \(t=\frac{x-1}{1-2 x}\) \(\text { So, } f(x)=\frac{x-1}{1-2 x}+1=\frac{x-1+1-2 x}{1-2 x}=\frac{x}{2 x-1}\) \(\int \frac{x}{2 x-1} d x=\frac{1}{2} \int \frac{2 x-1+1}{2 x-1} d x\) \(=\frac{1}{2}\left[\int \frac{2 x-1}{2 x-1} d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[\int d x+\int \frac{1}{2 x-1} d x\right]\) \(=\frac{1}{2}\left[x+\frac{1}{2} \log |2 x-1|\right]+C\) \(=\frac{x}{2}+\frac{1}{4} \log |2 x-1|+C\)
AP EAMCET-2018-24.04.2018
Integral Calculus
86147
\(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x=\)
1 \(\frac{x^{5}}{4}+x^{3}+6 x^{2}+c\)
2 \(\frac{x^{5}}{5}+\frac{x^{4}}{4}+6 x+c\)
3 \(\frac{x^{5}}{5}+x^{3}+6 x+c\)
4 \(\frac{x^{5}}{5}-\frac{x^{3}}{2}+6 x^{2}+c\)
Explanation:
(C) : Given, \(\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x\) \(x^{8}-9 x^{2}+18\) is divided by \(x^{4}-3 x^{2}+3\), we get quotient \(\mathrm{x}^{4}+3 \mathrm{x}^{2}+6\) and 0 remainder. Now, \(\quad \int\left(x^{4}+3 x^{2}+6\right) d x\) \(=\frac{x^{5}}{5}+\frac{3 x^{3}}{3}+6 x+C=\frac{x^{5}}{5}+x^{3}+6 x+C\)
