Question

$\color{blue} \displaystyle \sf \int \bigg( \sum_{n = 1} ^{ \infty } \bigg( \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \frac{1}{(x + {h}^{2} )} - \frac{1}{ {x}^{2} } }{h} \bigg) \bigg) \bigg)^{n} \bigg) \: dx$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \frac{1}{ \left(x +h \right)^{2} } - \frac{1}{ {x}^{2} } }{h} \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ {x}^{2} - \left(x + h \right)^{2} }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \left(x - x - h \right) \left( x + x + h\right) }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ - h \left( 2 x + h\right) }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ - \left( 2 x + h\right) }{ {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \bigg( \frac{ - 2 x }{ {x}^{2} \cdot {x}^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \frac{ - 2 }{ {x}^{3} } \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{ (- 2)( - 3) }{ {x}^{4} }\right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[6 \sum_{n = 1} ^{ \infty } \frac{ 1 }{ {x}^{4n} }\right] dx$

$\displaystyle = \sf6 \int \left [ \frac{ 1 }{ {x}^{4} } +\frac{ 1 }{ {x}^{8} } +\frac{ 1 }{ {x}^{12} } +\frac{ 1 }{ {x}^{16} } + \cdots\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ 1 +\frac{ 1 }{ {x}^{4}} +\frac{ 1 }{ {x}^{8} } +\frac{ 1 }{ {x}^{12} } + \cdots\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ \dfrac{1 }{1 -\dfrac{ 1 }{ {x}^{4}} }\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ \dfrac{ {x}^{4} }{{x}^{4} - 1 }\right] dx$

$\displaystyle = \sf6 \int \dfrac{ 1 }{{x}^{4} - 1 }\ dx$

$\displaystyle = \sf6 \int \dfrac{ 1 }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ 2 }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ \left({x}^{2} + 1 \right) -\left({x}^{2} - 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ \left({x}^{2} + 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) } dx - 3 \int \dfrac{ \left({x}^{2} - 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) } dx$

$\displaystyle = \sf3 \int \dfrac{ dx }{ {x}^{2} - 1 } - 3 \int \dfrac{ dx }{ {x}^{2} + 1 }$

$\displaystyle = \sf3 \cdot \dfrac{1}{2}\ln \left| \dfrac{ x - 1 }{ x + 1 } \right| - 3 { tan}^{ - 1}(x) +C$

$\displaystyle = \sf3\ln \left( \sqrt{ \dfrac{ x - 1 }{ x + 1 }} \right) - 3 { tan}^{ - 1}(x) +C$

Answer 2

Answer:

Step-by-step explanation:

We have,

$\displaystyle \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \frac{1}{ \left(x +h \right)^{2} } - \frac{1}{ {x}^{2} } }{h} \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ {x}^{2} - \left(x + h \right)^{2} }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \left(x - x - h \right) \left( x + x + h\right) }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ - h \left( 2 x + h\right) }{h {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ - \left( 2 x + h\right) }{ {x}^{2} \left(x + h \right)^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \bigg( \frac{ - 2 x }{ {x}^{2} \cdot {x}^{2} } \bigg) \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{d}{dx} \bigg( \frac{ - 2 }{ {x}^{3} } \bigg) \right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[ \sum_{n = 1} ^{ \infty } \left \{ \frac{ (- 2)( - 3) }{ {x}^{4} }\right \}^{n} \right] dx$

$\displaystyle = \sf \int \left[6 \sum_{n = 1} ^{ \infty } \frac{ 1 }{ {x}^{4n} }\right] dx$

$\displaystyle = \sf6 \int \left [ \frac{ 1 }{ {x}^{4} } +\frac{ 1 }{ {x}^{8} } +\frac{ 1 }{ {x}^{12} } +\frac{ 1 }{ {x}^{16} } + \cdots\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ 1 +\frac{ 1 }{ {x}^{4}} +\frac{ 1 }{ {x}^{8} } +\frac{ 1 }{ {x}^{12} } + \cdots\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ \dfrac{1 }{1 -\dfrac{ 1 }{ {x}^{4}} }\right] dx$

$\displaystyle = \sf6 \int\frac{ 1 }{ {x}^{4} } \left [ \dfrac{ {x}^{4} }{{x}^{4} - 1 }\right] dx$

$\displaystyle = \sf6 \int \dfrac{ 1 }{{x}^{4} - 1 }\ dx$

$\displaystyle = \sf6 \int \dfrac{ 1 }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ 2 }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ \left({x}^{2} + 1 \right) -\left({x}^{2} - 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) }\ dx$

$\displaystyle = \sf3 \int \dfrac{ \left({x}^{2} + 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) } dx - 3 \int \dfrac{ \left({x}^{2} - 1 \right) }{ \left({x}^{2} - 1 \right) \left({x}^{2} + 1 \right) } dx$

$\displaystyle = \sf3 \int \dfrac{ dx }{ {x}^{2} - 1 } - 3 \int \dfrac{ dx }{ {x}^{2} + 1 }$

$\displaystyle = \sf3 \cdot \dfrac{1}{2}\ln \left| \dfrac{ x - 1 }{ x + 1 } \right| - 3 { tan}^{ - 1}(x) +C$

$\displaystyle = \sf3\ln \left( \sqrt{ \dfrac{ x - 1 }{ x + 1 }} \right) - 3 { tan}^{ - 1}(x) +C$