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∫{1+(logx)2(logx−1)2}dx is equal to
1+x2xex+C
(logx)2+1x+C
(Correct Answer)(logx)2+1logx+C
x2+1x+C
(logx)2+1x+C
Concept:
[ex[f(x)+f′(x)]dx=exf(x)+c
\begin{aligned}
Calculation:
& \mathbf{I}=\int\left({\frac{\left(\log\mathbf{x}-1\right)}{1+\left(\log\mathbf{x}\right)^{2}}}\right)^{2}\mathrm{dx} \\
& \mathrm{Letlogx=t\Leftrightarrow x=e^{t}} \\
& \text{Differentiating with respect to x, we get} \\
& \Rightarrow\frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{dt} \\
& \Rightarrow\mathrm{dx=xdt} \\
& \Rightarrow\mathrm{dx}=\mathrm{e^{t}dt} \\
& \mathrm{Now}, \\
& \mathbf{I}=\int\left({\frac{(\mathbf{t}-1)}{1+\mathbf{t}^{2}}}\right)^{2}\mathbf{e}^{\mathbf{t}}\mathrm{dt} \\
& =\int\frac{(\mathrm{t^{2}+1-2t})}{(1+\mathrm{t^{2}})^{2}}\mathrm{e^{t}dt} \\
& =\int\left[\frac{\mathrm{t}^{2}+1}{\left(\mathrm{t}^{2}+1\right)^{2}}-\frac{2\mathrm{t}}{\left(\mathrm{t}^{2}+1\right)^{2}}\right]\mathrm{e}^{\mathrm{t}}\mathrm{dt} \\
& =\int\mathrm{e}^{\mathrm{t}}\left[\frac{1}{\left(\mathrm{t}^{2}+1\right)}-\frac{2\mathrm{t}}{\left(\mathrm{t}^{2}+1\right)^{2}}\right]\mathrm{dt} \\
& \mathrm{Let~f(t)}={\frac{1}{(t^{2}+1)}}
\end{aligned}
Differentiating with respect to t, we get
⇒f′(t)=(t2+1)2−2t
I = ∫e^t[f'(t)+f(t)]\,dt
= etf(t)+c
= et(t2+1)1+c
Resubstitute the value of t and et, we get
I = (logx)2+1x+C
Concept:
[ex[f(x)+f′(x)]dx=exf(x)+c
\begin{aligned}
Calculation:
& \mathbf{I}=\int\left({\frac{\left(\log\mathbf{x}-1\right)}{1+\left(\log\mathbf{x}\right)^{2}}}\right)^{2}\mathrm{dx} \\
& \mathrm{Letlogx=t\Leftrightarrow x=e^{t}} \\
& \text{Differentiating with respect to x, we get} \\
& \Rightarrow\frac{1}{\mathrm{x}}\mathrm{dx}=\mathrm{dt} \\
& \Rightarrow\mathrm{dx=xdt} \\
& \Rightarrow\mathrm{dx}=\mathrm{e^{t}dt} \\
& \mathrm{Now}, \\
& \mathbf{I}=\int\left({\frac{(\mathbf{t}-1)}{1+\mathbf{t}^{2}}}\right)^{2}\mathbf{e}^{\mathbf{t}}\mathrm{dt} \\
& =\int\frac{(\mathrm{t^{2}+1-2t})}{(1+\mathrm{t^{2}})^{2}}\mathrm{e^{t}dt} \\
& =\int\left[\frac{\mathrm{t}^{2}+1}{\left(\mathrm{t}^{2}+1\right)^{2}}-\frac{2\mathrm{t}}{\left(\mathrm{t}^{2}+1\right)^{2}}\right]\mathrm{e}^{\mathrm{t}}\mathrm{dt} \\
& =\int\mathrm{e}^{\mathrm{t}}\left[\frac{1}{\left(\mathrm{t}^{2}+1\right)}-\frac{2\mathrm{t}}{\left(\mathrm{t}^{2}+1\right)^{2}}\right]\mathrm{dt} \\
& \mathrm{Let~f(t)}={\frac{1}{(t^{2}+1)}}
\end{aligned}
Differentiating with respect to t, we get
⇒f′(t)=(t2+1)2−2t
I = ∫e^t[f'(t)+f(t)]\,dt
= etf(t)+c
= et(t2+1)1+c
Resubstitute the value of t and et, we get
I = (logx)2+1x+C