(B) : Given, \(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x \quad \because\left(e^{\log _{e} x}=x\right)\) \(\mathrm{e}^{\log x^{3}}\left(\mathrm{x}^{4}+1\right)^{-1}=\frac{\mathrm{x}^{3}}{\mathrm{x}^{4}+1}\) Let, \(\mathrm{x}^{4}+1=\mathrm{t}\) \(4 x^{3} d x=d t\) So, \(=\int \mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1} \mathrm{dx}\) \(=\int \frac{x^{3}}{x^{4}+1} d x=\frac{1}{4} \int \frac{d t}{t}\) \(=\frac{1}{4} \log |\mathrm{t}|+\mathrm{C}=\frac{1}{4} \log \left|\mathrm{x}^{4}+1\right|+\mathrm{C}\) \(=\frac{1}{4} \log \left(\mathrm{x}^{4}+1\right)+\mathrm{C}\)
AP EAMCET-2020-22.09.2020
Integral Calculus
86322
If \(\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+C\), where \(C\) is a constant of integration then \(\mathrm{g}(-1)\) is equal to
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(-\frac{5}{2}\)
Explanation:
(D) : Given, \(\int \mathrm{x}^{5} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{-\mathrm{x}^{2}}+\mathrm{C}\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(=\frac{1}{2} \int \mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}\) \(=\frac{1}{2}\left[-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t}+\mathrm{e}^{-\mathrm{t}} \mathrm{dt}\right]\) \(=\frac{-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}}{2}-\mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\) \(=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C\) Now comparing with the equation \(\mathrm{g}(\mathrm{x})=-\frac{1}{2}\left(\mathrm{x}^{4}+2 \mathrm{x}^{2}+2\right)\) \(g(-1)=-\frac{1}{2}(1+2+2)=\frac{-5}{2}\)
JEE Main-2019-10.04.2019
Integral Calculus
86323
If \(\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C\) where, \(C\) is a constant of integration, then the function \(f(x)\) is equal to
(B) : Given, \(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x \quad \because\left(e^{\log _{e} x}=x\right)\) \(\mathrm{e}^{\log x^{3}}\left(\mathrm{x}^{4}+1\right)^{-1}=\frac{\mathrm{x}^{3}}{\mathrm{x}^{4}+1}\) Let, \(\mathrm{x}^{4}+1=\mathrm{t}\) \(4 x^{3} d x=d t\) So, \(=\int \mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1} \mathrm{dx}\) \(=\int \frac{x^{3}}{x^{4}+1} d x=\frac{1}{4} \int \frac{d t}{t}\) \(=\frac{1}{4} \log |\mathrm{t}|+\mathrm{C}=\frac{1}{4} \log \left|\mathrm{x}^{4}+1\right|+\mathrm{C}\) \(=\frac{1}{4} \log \left(\mathrm{x}^{4}+1\right)+\mathrm{C}\)
AP EAMCET-2020-22.09.2020
Integral Calculus
86322
If \(\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+C\), where \(C\) is a constant of integration then \(\mathrm{g}(-1)\) is equal to
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(-\frac{5}{2}\)
Explanation:
(D) : Given, \(\int \mathrm{x}^{5} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{-\mathrm{x}^{2}}+\mathrm{C}\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(=\frac{1}{2} \int \mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}\) \(=\frac{1}{2}\left[-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t}+\mathrm{e}^{-\mathrm{t}} \mathrm{dt}\right]\) \(=\frac{-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}}{2}-\mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\) \(=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C\) Now comparing with the equation \(\mathrm{g}(\mathrm{x})=-\frac{1}{2}\left(\mathrm{x}^{4}+2 \mathrm{x}^{2}+2\right)\) \(g(-1)=-\frac{1}{2}(1+2+2)=\frac{-5}{2}\)
JEE Main-2019-10.04.2019
Integral Calculus
86323
If \(\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C\) where, \(C\) is a constant of integration, then the function \(f(x)\) is equal to
(B) : Given, \(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x \quad \because\left(e^{\log _{e} x}=x\right)\) \(\mathrm{e}^{\log x^{3}}\left(\mathrm{x}^{4}+1\right)^{-1}=\frac{\mathrm{x}^{3}}{\mathrm{x}^{4}+1}\) Let, \(\mathrm{x}^{4}+1=\mathrm{t}\) \(4 x^{3} d x=d t\) So, \(=\int \mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1} \mathrm{dx}\) \(=\int \frac{x^{3}}{x^{4}+1} d x=\frac{1}{4} \int \frac{d t}{t}\) \(=\frac{1}{4} \log |\mathrm{t}|+\mathrm{C}=\frac{1}{4} \log \left|\mathrm{x}^{4}+1\right|+\mathrm{C}\) \(=\frac{1}{4} \log \left(\mathrm{x}^{4}+1\right)+\mathrm{C}\)
AP EAMCET-2020-22.09.2020
Integral Calculus
86322
If \(\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+C\), where \(C\) is a constant of integration then \(\mathrm{g}(-1)\) is equal to
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(-\frac{5}{2}\)
Explanation:
(D) : Given, \(\int \mathrm{x}^{5} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{-\mathrm{x}^{2}}+\mathrm{C}\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(=\frac{1}{2} \int \mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}\) \(=\frac{1}{2}\left[-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t}+\mathrm{e}^{-\mathrm{t}} \mathrm{dt}\right]\) \(=\frac{-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}}{2}-\mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\) \(=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C\) Now comparing with the equation \(\mathrm{g}(\mathrm{x})=-\frac{1}{2}\left(\mathrm{x}^{4}+2 \mathrm{x}^{2}+2\right)\) \(g(-1)=-\frac{1}{2}(1+2+2)=\frac{-5}{2}\)
JEE Main-2019-10.04.2019
Integral Calculus
86323
If \(\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C\) where, \(C\) is a constant of integration, then the function \(f(x)\) is equal to
(B) : Given, \(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x \quad \because\left(e^{\log _{e} x}=x\right)\) \(\mathrm{e}^{\log x^{3}}\left(\mathrm{x}^{4}+1\right)^{-1}=\frac{\mathrm{x}^{3}}{\mathrm{x}^{4}+1}\) Let, \(\mathrm{x}^{4}+1=\mathrm{t}\) \(4 x^{3} d x=d t\) So, \(=\int \mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1} \mathrm{dx}\) \(=\int \frac{x^{3}}{x^{4}+1} d x=\frac{1}{4} \int \frac{d t}{t}\) \(=\frac{1}{4} \log |\mathrm{t}|+\mathrm{C}=\frac{1}{4} \log \left|\mathrm{x}^{4}+1\right|+\mathrm{C}\) \(=\frac{1}{4} \log \left(\mathrm{x}^{4}+1\right)+\mathrm{C}\)
AP EAMCET-2020-22.09.2020
Integral Calculus
86322
If \(\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+C\), where \(C\) is a constant of integration then \(\mathrm{g}(-1)\) is equal to
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(-\frac{5}{2}\)
Explanation:
(D) : Given, \(\int \mathrm{x}^{5} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{-\mathrm{x}^{2}}+\mathrm{C}\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(=\frac{1}{2} \int \mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}\) \(=\frac{1}{2}\left[-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t}+\mathrm{e}^{-\mathrm{t}} \mathrm{dt}\right]\) \(=\frac{-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}}{2}-\mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\) \(=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C\) Now comparing with the equation \(\mathrm{g}(\mathrm{x})=-\frac{1}{2}\left(\mathrm{x}^{4}+2 \mathrm{x}^{2}+2\right)\) \(g(-1)=-\frac{1}{2}(1+2+2)=\frac{-5}{2}\)
JEE Main-2019-10.04.2019
Integral Calculus
86323
If \(\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C\) where, \(C\) is a constant of integration, then the function \(f(x)\) is equal to
(B) : Given, \(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x \quad \because\left(e^{\log _{e} x}=x\right)\) \(\mathrm{e}^{\log x^{3}}\left(\mathrm{x}^{4}+1\right)^{-1}=\frac{\mathrm{x}^{3}}{\mathrm{x}^{4}+1}\) Let, \(\mathrm{x}^{4}+1=\mathrm{t}\) \(4 x^{3} d x=d t\) So, \(=\int \mathrm{e}^{3 \log \mathrm{x}}\left(\mathrm{x}^{4}+1\right)^{-1} \mathrm{dx}\) \(=\int \frac{x^{3}}{x^{4}+1} d x=\frac{1}{4} \int \frac{d t}{t}\) \(=\frac{1}{4} \log |\mathrm{t}|+\mathrm{C}=\frac{1}{4} \log \left|\mathrm{x}^{4}+1\right|+\mathrm{C}\) \(=\frac{1}{4} \log \left(\mathrm{x}^{4}+1\right)+\mathrm{C}\)
AP EAMCET-2020-22.09.2020
Integral Calculus
86322
If \(\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+C\), where \(C\) is a constant of integration then \(\mathrm{g}(-1)\) is equal to
1 -1
2 1
3 \(-\frac{1}{2}\)
4 \(-\frac{5}{2}\)
Explanation:
(D) : Given, \(\int \mathrm{x}^{5} \mathrm{e}^{-\mathrm{x}^{2}} \mathrm{dx}=\mathrm{g}(\mathrm{x}) \mathrm{e}^{-\mathrm{x}^{2}}+\mathrm{C}\) Let, \(\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(=\frac{1}{2} \int \mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}\) \(=\frac{1}{2}\left[-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t}+\mathrm{e}^{-\mathrm{t}} \mathrm{dt}\right]\) \(=\frac{-\mathrm{t}^{2} \mathrm{e}^{-\mathrm{t}}}{2}-\mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}}-\mathrm{e}^{-\mathrm{t}}\) \(=\left(-\frac{x^{4}}{2}-x^{2}-1\right) e^{-x^{2}}+C\) Now comparing with the equation \(\mathrm{g}(\mathrm{x})=-\frac{1}{2}\left(\mathrm{x}^{4}+2 \mathrm{x}^{2}+2\right)\) \(g(-1)=-\frac{1}{2}(1+2+2)=\frac{-5}{2}\)
JEE Main-2019-10.04.2019
Integral Calculus
86323
If \(\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C\) where, \(C\) is a constant of integration, then the function \(f(x)\) is equal to
