NEET Test Series from KOTA - 10 Papers In MS WORD
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Integral Calculus
86319
The integral \(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x\) is equal to (where \(\mathbf{C}\) is a constant of integration)
(B) : Given, \(I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\int \frac{\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right)}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\text { Let, } 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=t \Rightarrow 0-\left(\frac{6}{x^{3}}+\frac{4}{x^{5}}\right)=\frac{d t}{d x}\) \(-2\left(\frac{3}{x^{2}}+\frac{2}{x^{5}}\right) d x=d t \Rightarrow\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=\frac{-1}{2} d t\) \(\therefore I=\frac{-1}{2} \int \frac{1}{t^{4}} d t \Rightarrow I=\frac{1}{6\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}\) \(I=\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C\)
JEE Main-2019-12.01.2019
Integral Calculus
86320
The integral \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) is equal to
(B) : Given, \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x d x}{\left\{\sin ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)+\cos ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)\right\}^{2}} d x\) \(=\int \frac{\sin ^{2} x \cos ^{2} x}{\left\{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{3} x+\cos ^{3}\right)\right\}^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x\) On dividing by \(\cos ^{3} \mathrm{x}\) in nominator for and denominator we get \(\int \frac{\sec ^{2} x \tan ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x\) Let, \(1+\tan ^{3} \mathrm{x}=\mathrm{v}\) \(\frac{d v}{d x}=3 \tan ^{2} x \sec ^{2} x d x\) \(=\frac{1}{3} \int \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-\frac{1}{3} \frac{1}{\mathrm{v}}+\mathrm{C}=-\frac{1}{3\left(1+\tan ^{3} \mathrm{x}\right)}+\mathrm{C}\)
JEE Main-2018
Integral Calculus
86321
If \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\), where \(C\) is a constant of integration, then \(f(x)\) is equal to
1 \(-4 x^{3}-1\)
2 \(4 \mathrm{x}^{3}+1\)
3 \(-2 \mathrm{x}^{3}-1\)
4 \(-2 \mathrm{x}^{3}+1\)
Explanation:
(A) : Given, \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\) Let, \(\mathrm{x}^{3}=\mathrm{v}\) \(3 \mathrm{x}^{2} \cdot \mathrm{dx}=\mathrm{dv}\) \(\int x^{3} e^{-4 x^{3}} x^{2} d x=\frac{1}{3} \int v e^{-4 v} d v\) \(=\frac{1}{3}\left[\mathrm{v} \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4}-\int \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4} \mathrm{dv}\right]=\frac{-\mathrm{e}^{-4 \mathrm{v}}}{48}[4 \mathrm{v}+1]+\mathrm{C}\) \(=-\frac{\mathrm{e}^{-4 \mathrm{x}^{3}}}{48}\left[4 \mathrm{x}^{3}+1\right]+\mathrm{C}\) \(\therefore \mathrm{f}(\mathrm{x})=-1-4 \mathrm{x}^{3}\)
JEE Main-2019-10.01.2019
Integral Calculus
86327
If \(\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|\) \(+C\) where \(C\) is a constant of integration, then the ordered pair \((\lambda, f(\theta))\) is equal to
86319
The integral \(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x\) is equal to (where \(\mathbf{C}\) is a constant of integration)
(B) : Given, \(I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\int \frac{\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right)}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\text { Let, } 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=t \Rightarrow 0-\left(\frac{6}{x^{3}}+\frac{4}{x^{5}}\right)=\frac{d t}{d x}\) \(-2\left(\frac{3}{x^{2}}+\frac{2}{x^{5}}\right) d x=d t \Rightarrow\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=\frac{-1}{2} d t\) \(\therefore I=\frac{-1}{2} \int \frac{1}{t^{4}} d t \Rightarrow I=\frac{1}{6\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}\) \(I=\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C\)
JEE Main-2019-12.01.2019
Integral Calculus
86320
The integral \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) is equal to
(B) : Given, \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x d x}{\left\{\sin ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)+\cos ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)\right\}^{2}} d x\) \(=\int \frac{\sin ^{2} x \cos ^{2} x}{\left\{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{3} x+\cos ^{3}\right)\right\}^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x\) On dividing by \(\cos ^{3} \mathrm{x}\) in nominator for and denominator we get \(\int \frac{\sec ^{2} x \tan ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x\) Let, \(1+\tan ^{3} \mathrm{x}=\mathrm{v}\) \(\frac{d v}{d x}=3 \tan ^{2} x \sec ^{2} x d x\) \(=\frac{1}{3} \int \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-\frac{1}{3} \frac{1}{\mathrm{v}}+\mathrm{C}=-\frac{1}{3\left(1+\tan ^{3} \mathrm{x}\right)}+\mathrm{C}\)
JEE Main-2018
Integral Calculus
86321
If \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\), where \(C\) is a constant of integration, then \(f(x)\) is equal to
1 \(-4 x^{3}-1\)
2 \(4 \mathrm{x}^{3}+1\)
3 \(-2 \mathrm{x}^{3}-1\)
4 \(-2 \mathrm{x}^{3}+1\)
Explanation:
(A) : Given, \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\) Let, \(\mathrm{x}^{3}=\mathrm{v}\) \(3 \mathrm{x}^{2} \cdot \mathrm{dx}=\mathrm{dv}\) \(\int x^{3} e^{-4 x^{3}} x^{2} d x=\frac{1}{3} \int v e^{-4 v} d v\) \(=\frac{1}{3}\left[\mathrm{v} \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4}-\int \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4} \mathrm{dv}\right]=\frac{-\mathrm{e}^{-4 \mathrm{v}}}{48}[4 \mathrm{v}+1]+\mathrm{C}\) \(=-\frac{\mathrm{e}^{-4 \mathrm{x}^{3}}}{48}\left[4 \mathrm{x}^{3}+1\right]+\mathrm{C}\) \(\therefore \mathrm{f}(\mathrm{x})=-1-4 \mathrm{x}^{3}\)
JEE Main-2019-10.01.2019
Integral Calculus
86327
If \(\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|\) \(+C\) where \(C\) is a constant of integration, then the ordered pair \((\lambda, f(\theta))\) is equal to
86319
The integral \(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x\) is equal to (where \(\mathbf{C}\) is a constant of integration)
(B) : Given, \(I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\int \frac{\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right)}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\text { Let, } 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=t \Rightarrow 0-\left(\frac{6}{x^{3}}+\frac{4}{x^{5}}\right)=\frac{d t}{d x}\) \(-2\left(\frac{3}{x^{2}}+\frac{2}{x^{5}}\right) d x=d t \Rightarrow\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=\frac{-1}{2} d t\) \(\therefore I=\frac{-1}{2} \int \frac{1}{t^{4}} d t \Rightarrow I=\frac{1}{6\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}\) \(I=\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C\)
JEE Main-2019-12.01.2019
Integral Calculus
86320
The integral \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) is equal to
(B) : Given, \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x d x}{\left\{\sin ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)+\cos ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)\right\}^{2}} d x\) \(=\int \frac{\sin ^{2} x \cos ^{2} x}{\left\{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{3} x+\cos ^{3}\right)\right\}^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x\) On dividing by \(\cos ^{3} \mathrm{x}\) in nominator for and denominator we get \(\int \frac{\sec ^{2} x \tan ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x\) Let, \(1+\tan ^{3} \mathrm{x}=\mathrm{v}\) \(\frac{d v}{d x}=3 \tan ^{2} x \sec ^{2} x d x\) \(=\frac{1}{3} \int \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-\frac{1}{3} \frac{1}{\mathrm{v}}+\mathrm{C}=-\frac{1}{3\left(1+\tan ^{3} \mathrm{x}\right)}+\mathrm{C}\)
JEE Main-2018
Integral Calculus
86321
If \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\), where \(C\) is a constant of integration, then \(f(x)\) is equal to
1 \(-4 x^{3}-1\)
2 \(4 \mathrm{x}^{3}+1\)
3 \(-2 \mathrm{x}^{3}-1\)
4 \(-2 \mathrm{x}^{3}+1\)
Explanation:
(A) : Given, \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\) Let, \(\mathrm{x}^{3}=\mathrm{v}\) \(3 \mathrm{x}^{2} \cdot \mathrm{dx}=\mathrm{dv}\) \(\int x^{3} e^{-4 x^{3}} x^{2} d x=\frac{1}{3} \int v e^{-4 v} d v\) \(=\frac{1}{3}\left[\mathrm{v} \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4}-\int \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4} \mathrm{dv}\right]=\frac{-\mathrm{e}^{-4 \mathrm{v}}}{48}[4 \mathrm{v}+1]+\mathrm{C}\) \(=-\frac{\mathrm{e}^{-4 \mathrm{x}^{3}}}{48}\left[4 \mathrm{x}^{3}+1\right]+\mathrm{C}\) \(\therefore \mathrm{f}(\mathrm{x})=-1-4 \mathrm{x}^{3}\)
JEE Main-2019-10.01.2019
Integral Calculus
86327
If \(\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|\) \(+C\) where \(C\) is a constant of integration, then the ordered pair \((\lambda, f(\theta))\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86319
The integral \(\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x\) is equal to (where \(\mathbf{C}\) is a constant of integration)
(B) : Given, \(I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\int \frac{\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right)}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x\) \(\text { Let, } 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=t \Rightarrow 0-\left(\frac{6}{x^{3}}+\frac{4}{x^{5}}\right)=\frac{d t}{d x}\) \(-2\left(\frac{3}{x^{2}}+\frac{2}{x^{5}}\right) d x=d t \Rightarrow\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=\frac{-1}{2} d t\) \(\therefore I=\frac{-1}{2} \int \frac{1}{t^{4}} d t \Rightarrow I=\frac{1}{6\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}\) \(I=\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C\)
JEE Main-2019-12.01.2019
Integral Calculus
86320
The integral \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) is equal to
(B) : Given, \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{5} x+\cos ^{3} x \sin ^{2} x+\sin ^{3} x \cos ^{2} x+\cos ^{5} x\right)^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x d x}{\left\{\sin ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)+\cos ^{2} x\left(\sin ^{3} x+\cos ^{3} x\right)\right\}^{2}} d x\) \(=\int \frac{\sin ^{2} x \cos ^{2} x}{\left\{\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{3} x+\cos ^{3}\right)\right\}^{2}} d x\) \(\int \frac{\sin ^{2} x \cos ^{2} x}{\left(\sin ^{3} x+\cos ^{3} x\right)^{2}} d x\) On dividing by \(\cos ^{3} \mathrm{x}\) in nominator for and denominator we get \(\int \frac{\sec ^{2} x \tan ^{2} x}{\left(\tan ^{3} x+1\right)^{2}} d x\) Let, \(1+\tan ^{3} \mathrm{x}=\mathrm{v}\) \(\frac{d v}{d x}=3 \tan ^{2} x \sec ^{2} x d x\) \(=\frac{1}{3} \int \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-\frac{1}{3} \frac{1}{\mathrm{v}}+\mathrm{C}=-\frac{1}{3\left(1+\tan ^{3} \mathrm{x}\right)}+\mathrm{C}\)
JEE Main-2018
Integral Calculus
86321
If \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\), where \(C\) is a constant of integration, then \(f(x)\) is equal to
1 \(-4 x^{3}-1\)
2 \(4 \mathrm{x}^{3}+1\)
3 \(-2 \mathrm{x}^{3}-1\)
4 \(-2 \mathrm{x}^{3}+1\)
Explanation:
(A) : Given, \(\int x^{5} e^{-4 x^{3}} d x=\frac{1}{48} e^{-4 x^{3}} f(x)+C\) Let, \(\mathrm{x}^{3}=\mathrm{v}\) \(3 \mathrm{x}^{2} \cdot \mathrm{dx}=\mathrm{dv}\) \(\int x^{3} e^{-4 x^{3}} x^{2} d x=\frac{1}{3} \int v e^{-4 v} d v\) \(=\frac{1}{3}\left[\mathrm{v} \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4}-\int \frac{\mathrm{e}^{-4 \mathrm{v}}}{-4} \mathrm{dv}\right]=\frac{-\mathrm{e}^{-4 \mathrm{v}}}{48}[4 \mathrm{v}+1]+\mathrm{C}\) \(=-\frac{\mathrm{e}^{-4 \mathrm{x}^{3}}}{48}\left[4 \mathrm{x}^{3}+1\right]+\mathrm{C}\) \(\therefore \mathrm{f}(\mathrm{x})=-1-4 \mathrm{x}^{3}\)
JEE Main-2019-10.01.2019
Integral Calculus
86327
If \(\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|\) \(+C\) where \(C\) is a constant of integration, then the ordered pair \((\lambda, f(\theta))\) is equal to