कोरूना चा वाढता प्रादुर्भाव थांबविण्यासाठी तुम्ही कोणत्या उपाययोजना कराल
Answers
Explanation:
Solution−
Given integral is
\begin{gathered}\rm \: \displaystyle\int\rm \bigg(log(logx) + \dfrac{1}{ {(logx)}^{2} } \bigg) \: dx \\ \end{gathered}
∫(log(logx)+
(logx)
2
1
)dx
To solve this integral, we use method of substitution.
So, substituting
\begin{gathered}\rm \: logx = y \\ \end{gathered}
logx=y
\begin{gathered}\rm \: x = {e}^{y} \\ \end{gathered}
x=e
y
\begin{gathered}\rm \: dx = {e}^{y}dy \\ \end{gathered}
dx=e
y
dy
So, on substituting all these values, we get
\rm \: = \displaystyle\int\rm \bigg(logy + \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy=∫(logy+
y
2
1
)e
y
dy
\begin{gathered}\rm \: = \displaystyle\int\rm \bigg(logy + \dfrac{1}{y} - \dfrac{1}{y} + \dfrac{1}{ {y}^{2} } \bigg) {e}^{y} \: dy \\ \end{gathered}
=∫(logy+
y
1
−
y
1
+
y
2
1
)e
y
dy
can be further rewritten as
\begin{gathered}\rm \: = \displaystyle\int\rm \bigg(logy + \dfrac{1}{y} \bigg) {e}^{y}dy - \displaystyle\int\rm \bigg(\dfrac{1}{y} - \dfrac{1}{ {y}^{2} } \bigg) {e}^{y}dy \\ \end{gathered}
=∫(logy+
y
1
)e
y
dy−∫(
y
1
−
y
2
1
)e
y
dy
We know,
\begin{gathered}\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)]dx \: = \: {e}^{x}f(x) + c \: }} \\ \end{gathered}
∫e
x
[f(x)+f
′
(x)]dx=e
x
f(x)+c
So, for the first integral,
\begin{gathered}\rm \: f(y) = logy \\ \end{gathered}
f(y)=logy
\begin{gathered}\rm \: f'(y) = \dfrac{1}{y} \\ \end{gathered}
f
′
(y)=
y
1
And, for the second integral,
\begin{gathered}\rm \: f(y) = \dfrac{1}{y} \\ \end{gathered}
f(y)=
y
1
\begin{gathered}\rm \: f'(y) = - \dfrac{1}{ {y}^{2} } \: \\ \end{gathered}
f
′
(y)=−
y
2
1
So, using the above formula, we get
\begin{gathered}\rm \: = \: {e}^{y}logy \: - \: {e}^{y} \times \dfrac{1}{y} + c \\ \end{gathered}
=e
y
logy−e
y
×
y
1
+c
\begin{gathered}\rm \: = \: {e}^{y}\bigg(logy - \dfrac{1}{y} \bigg) + c \\ \end{gathered}
=e
y
(logy−
y
1
)+c
\begin{gathered}\rm \: = \: x\bigg(log(logx) - \dfrac{1}{logx} \bigg) + c \\ \end{gathered}
=x(log(logx)−
logx
1
)+c
Hence,
\begin{gathered}\rm \:\displaystyle\int\rm \bigg(log(logx) + \frac{1}{ {(logx)}^{2} } \bigg)dx= x\bigg(log(logx)-\dfrac{1}{logx} \bigg)+ c \\ \end{gathered}
∫(log(logx)+
(logx)
2
1
)dx=x(log(logx)−
logx
1
)+c
\rule{190pt}{2pt}
Remark :- Proof of the property
\begin{gathered}\boxed{ \rm{ \:\displaystyle\int\rm {e}^{x}[f(x) + f'(x)]dx \: = \: {e}^{x}f(x) + c \: }} \\ \end{gathered}
∫e
x
[f(x)+f
′
(x)]dx=e
x
f(x)+c
Consider,
\begin{gathered}\rm \: \displaystyle\int\rm {e}^{x}[f(x) + f'(x)] \: dx \\ \end{gathered}
∫e
x
[f(x)+f
′
(x)]dx
\begin{gathered}\rm \: = \displaystyle\int\rm {e}^{x}f(x)dx + \displaystyle\int\rm {e}^{x}f'(x) \: dx \\ \end{gathered}
=∫e
x
f(x)dx+∫e
x
f
′
(x)dx
Now, using integration by parts in first integral, we get
\begin{gathered}\rm \: = f(x)\displaystyle\int\rm {e}^{x}dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}f(x)\displaystyle\int\rm {e}^{x}dx \bigg]dx+ \displaystyle\int\rm {e}^{x}f'(x) \: dx \\ \end{gathered}
=f(x)∫e
x
dx−∫[
dx
d
f(x)∫e
x
dx]dx+∫e
x
f
′
(x)dx
\begin{gathered}\rm \: = {e}^{x}f(x) - \displaystyle\int\rm {e}^{x}f'(x) \: dx + \displaystyle\int\rm {e}^{x}f'(x) \: dx \: + \: c \\ \end{gathered}
=e
x
f(x)−∫e
x
f
′
(x)dx+∫e
x
f
′
(x)dx+c
\begin{gathered}\rm \: = \: {e}^{x}f(x)\: + \: c \\ \end{gathered}
=e
x
f(x)+c
\rule{190pt}{2pt}
Additional Information :-
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}
f(x)
k
sinx
cosx
sec
2
x
cosec
2
x
secxtanx
cosecxcotx
tanx
x
1
e
x
∫f(x)dx
kx+c
−cosx+c
sinx+c
tanx+c
−cotx+c
secx+c
−cosecx+c
logsecx+c
logx+c
e
x
+c