(A): Given, \(I=\int \frac{e^{x}(x+3)}{(x+5)^{3}} d x\) \(I=\int \frac{e^{x}(x+5-2)}{(x+5)^{3}} d x\) \(I=\int e^{x}\left(\frac{1}{(x+5)^{2}}-\frac{2}{(x+5)^{3}}\right) d x\) Let, \(\quad f(x)=\frac{1}{(x+5)^{2}}\) \(f^{\prime}(x)=\frac{-2}{(x+5)^{3}}\) So, \(I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) As we know, \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}+5)^{2}}+\mathrm{c}\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86301
If \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\), then \(g(x)\) is equal to
(B): Given, \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\) Let, \(I=\int \frac{x^{2}+1-2 x}{\left(x^{2}+1\right)^{2}} d x\) \(I=\int\left(\frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}}\right) d x\) Let, \(x^{2}+1=t\) \(2 x d x=d t\) \(I=\tan ^{-1}(x)-\int \frac{d t}{t^{2}}+k\) \(I=\tan ^{-1}(x)+\frac{1}{t}+k\) \(I=\tan ^{-1}(x)+\frac{1}{x^{2}+1}+k=\tan ^{-1}(x)+g(x)+k\) On comparing both side, we get - \(g(x)=\frac{1}{x^{2}+1}\)
(A): Given, \(I=\int \frac{e^{x}(x+3)}{(x+5)^{3}} d x\) \(I=\int \frac{e^{x}(x+5-2)}{(x+5)^{3}} d x\) \(I=\int e^{x}\left(\frac{1}{(x+5)^{2}}-\frac{2}{(x+5)^{3}}\right) d x\) Let, \(\quad f(x)=\frac{1}{(x+5)^{2}}\) \(f^{\prime}(x)=\frac{-2}{(x+5)^{3}}\) So, \(I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) As we know, \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}+5)^{2}}+\mathrm{c}\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86301
If \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\), then \(g(x)\) is equal to
(B): Given, \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\) Let, \(I=\int \frac{x^{2}+1-2 x}{\left(x^{2}+1\right)^{2}} d x\) \(I=\int\left(\frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}}\right) d x\) Let, \(x^{2}+1=t\) \(2 x d x=d t\) \(I=\tan ^{-1}(x)-\int \frac{d t}{t^{2}}+k\) \(I=\tan ^{-1}(x)+\frac{1}{t}+k\) \(I=\tan ^{-1}(x)+\frac{1}{x^{2}+1}+k=\tan ^{-1}(x)+g(x)+k\) On comparing both side, we get - \(g(x)=\frac{1}{x^{2}+1}\)
(A): Given, \(I=\int \frac{e^{x}(x+3)}{(x+5)^{3}} d x\) \(I=\int \frac{e^{x}(x+5-2)}{(x+5)^{3}} d x\) \(I=\int e^{x}\left(\frac{1}{(x+5)^{2}}-\frac{2}{(x+5)^{3}}\right) d x\) Let, \(\quad f(x)=\frac{1}{(x+5)^{2}}\) \(f^{\prime}(x)=\frac{-2}{(x+5)^{3}}\) So, \(I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) As we know, \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}+5)^{2}}+\mathrm{c}\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86301
If \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\), then \(g(x)\) is equal to
(B): Given, \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\) Let, \(I=\int \frac{x^{2}+1-2 x}{\left(x^{2}+1\right)^{2}} d x\) \(I=\int\left(\frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}}\right) d x\) Let, \(x^{2}+1=t\) \(2 x d x=d t\) \(I=\tan ^{-1}(x)-\int \frac{d t}{t^{2}}+k\) \(I=\tan ^{-1}(x)+\frac{1}{t}+k\) \(I=\tan ^{-1}(x)+\frac{1}{x^{2}+1}+k=\tan ^{-1}(x)+g(x)+k\) On comparing both side, we get - \(g(x)=\frac{1}{x^{2}+1}\)
(A): Given, \(I=\int \frac{e^{x}(x+3)}{(x+5)^{3}} d x\) \(I=\int \frac{e^{x}(x+5-2)}{(x+5)^{3}} d x\) \(I=\int e^{x}\left(\frac{1}{(x+5)^{2}}-\frac{2}{(x+5)^{3}}\right) d x\) Let, \(\quad f(x)=\frac{1}{(x+5)^{2}}\) \(f^{\prime}(x)=\frac{-2}{(x+5)^{3}}\) So, \(I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) As we know, \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}+5)^{2}}+\mathrm{c}\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86301
If \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\), then \(g(x)\) is equal to
(B): Given, \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\) Let, \(I=\int \frac{x^{2}+1-2 x}{\left(x^{2}+1\right)^{2}} d x\) \(I=\int\left(\frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}}\right) d x\) Let, \(x^{2}+1=t\) \(2 x d x=d t\) \(I=\tan ^{-1}(x)-\int \frac{d t}{t^{2}}+k\) \(I=\tan ^{-1}(x)+\frac{1}{t}+k\) \(I=\tan ^{-1}(x)+\frac{1}{x^{2}+1}+k=\tan ^{-1}(x)+g(x)+k\) On comparing both side, we get - \(g(x)=\frac{1}{x^{2}+1}\)
(A): Given, \(I=\int \frac{e^{x}(x+3)}{(x+5)^{3}} d x\) \(I=\int \frac{e^{x}(x+5-2)}{(x+5)^{3}} d x\) \(I=\int e^{x}\left(\frac{1}{(x+5)^{2}}-\frac{2}{(x+5)^{3}}\right) d x\) Let, \(\quad f(x)=\frac{1}{(x+5)^{2}}\) \(f^{\prime}(x)=\frac{-2}{(x+5)^{3}}\) So, \(I=\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x\) As we know, \(\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \frac{1}{(\mathrm{x}+5)^{2}}+\mathrm{c}\)
AP EAMCET-2021-19.08.2021
Integral Calculus
86301
If \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\), then \(g(x)\) is equal to
(B): Given, \(\int \frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}} d x=\tan ^{-1}(x)+g(x)+k\) Let, \(I=\int \frac{x^{2}+1-2 x}{\left(x^{2}+1\right)^{2}} d x\) \(I=\int\left(\frac{1}{x^{2}+1}-\frac{2 x}{\left(x^{2}+1\right)^{2}}\right) d x\) Let, \(x^{2}+1=t\) \(2 x d x=d t\) \(I=\tan ^{-1}(x)-\int \frac{d t}{t^{2}}+k\) \(I=\tan ^{-1}(x)+\frac{1}{t}+k\) \(I=\tan ^{-1}(x)+\frac{1}{x^{2}+1}+k=\tan ^{-1}(x)+g(x)+k\) On comparing both side, we get - \(g(x)=\frac{1}{x^{2}+1}\)
