Question

$\rm If \: f(x) = \int \frac{5 {x}^{8} + 7 {x}^{6} }{( {x {}^{2} + 1 + 2 {x}^{7} {)}}^{2} } \: dx \\ \\ \rm (x \geqslant 0) \: and \: f(0) = 0$
$\rm Then \: Find \: the \: value \: of \: f(1) \: .$
✈ Hoping For a great Answer.
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Answer 1

EXPLANATION.

$\sf \implies \displaystyle f(x) = \int \dfrac{5x^{8} + 7x^{6} }{(x^{2} + 1 + 2x^{7})^{2} } dx$

As we know that,

We can write equation as,

$\sf \implies \displaystyle f(x) = \int \dfrac{5x^{8} + 7x^{6} }{x^{14}(x^{-5} + x^{-7} + 2)^{2} } dx$

$\sf \implies \displaystyle f(x) = \int \dfrac{\bigg(\dfrac{5x^{8} }{x^{14} } + \dfrac{7x^{6} }{x^{14} } \bigg)}{(x^{-5}+ x^{-7} + 2)^{2} } dx$

$\sf \implies \displaystyle f(x) = \int \dfrac{5x^{-6} + 7x^{-8} }{(x^{-5} + x^{-7} + 2)^{2} } dx$

As we know that,

Now, we can apply substitution method in this equation, we get.

Let we assume that,

⇒ (x⁻⁵ + x⁻⁷ + 2) = t.

Differentiate w.r.t x, we get.

⇒ [-5x⁻⁶ + (-7x⁻⁸) + 0]dx = dt.

⇒ (-5x⁻⁶ - 7x⁻⁸)dx = dt.

⇒ (5x⁶ + 7x⁻⁸)dx = - dt.

Put the values in the equation, we get.

$\sf \implies \displaystyle f(x) =- \int \dfrac{dt}{t^{2} }$

$\sf \implies \displaystyle f(x) = - \int t^{-2} dt$

$\sf \implies \displaystyle f(x) = - \bigg[ \dfrac{t^{-2 + 1} }{-2 + 1} \bigg] + C.$

$\sf \implies \displaystyle f(x) = - \bigg[ \dfrac{t^{-1} }{- 1} \bigg] + C$

$\sf \implies \displaystyle f(x) = (t)^{-1} + C.$

$\sf \implies \displaystyle f(x) = \dfrac{1}{t} + C$

Put the value of t = (x⁻⁵ + x⁻⁷ + 2) in the equation, we get.

$\sf \implies \displaystyle f(x) = \dfrac{1}{(x^{-5} + x^{-7} + 2)} + C.$

Divide denominator by x⁻⁷ in the equation, we get.

$\sf \implies \displaystyle f(x) = \dfrac{1}{\bigg( \dfrac{x^{-5} }{x^{-7}}+ \dfrac{x^{-7} }{x^{-7} } + \dfrac{2}{x^{-7} }\bigg) } + C.$

$\sf \implies \displaystyle f(x) = \dfrac{x^{7} }{(x^{2} + 1 + 2x^{7} )} + C.$

Now, we put the value of f(0) = 0 in the equation, we get.

$\sf \implies \displaystyle f(0) = \dfrac{(0)^{7} }{[(0)^{2} + 1 + 2(0)^{7} ]} = 0$

Now, we put the value of f(1) in the equation, we get.

$\sf \implies \displaystyle f(1) = \dfrac{(1)^{7} }{[(1)^{2} + 1 + 2(1)^{7}] }$

$\sf \implies \displaystyle f(1) = \dfrac{1}{1 + 1 + 2} = \dfrac{1}{4}$

Answer 2

EXPLANATION.

Rightarrow f(x)= int 5x^ 8 +7x^ 6 (x^ 2 +1+2x^ 7 )^ 2 dx

As we know that,

We can write equation as,

Rightarrow f(x)= int 5x^ 8 +7x^ 6 x^ 14 (x^ -5 +x^ -7 +2)^ 2 dx Rightarrow f(x)= int ( 5x^ 8 x^ 14 + 7x^ 6 x^ 14 ) (x^ -5 +x^ -7 +2)^ 2 dx

Rightarrow f(x)= int 5x^ -6 +7x^ -8 (x^ -5 +x^ -7 +2)^ 2 dx

As we know that,

Now, we can apply substitution method in this equation, we get.

Let we assume that,

Rightarrow(x^ -5 +x^ -7 +2)=t.

Differentiate w.r.t x, we get.

Rightarrow[-5x^ -6 +(-7x^ -8 )+0]dx=dt.

Rightarrow(-5x^ -6 -7x^ -8 )dx=dt .

Rightarrow(5x^ 6 +7x^ -8 )dx=-dt.

Put the values in the equation, we get.

dt f(x) = - [ d

⇒ f(x)=

Rightarrow f(x)=-[ t^ -2+1 -2+1 ]+C

Rightarrow f(x)=-[ t^ -1 -1 ]+C

Rightarrow f(x)=(t)^ -1 +C.

Rightarrow f(x)= 1 t +C

Put the value of t = (x ^ - 5 + x ^ - 7 + 2) in the equation, we get.

Rightarrow f(x)= 1 (x^ -5 +x^ -7 +2) +C

Divide denominator by x-7 in the equation, we get.

Rightarrow f(x)= 1 ( x^ -5 x^ -7 + x^ -7 x^ -7 + 2 x^ -7 ) +C.

Rightarrow f(x)= x^ 7 (x^ 2 +1+2x^ 7 ) +C

Now, we put the value of f(0) = 0 in the

equation, we get.

⇒ f (0) (0)7 [(0)² + 1 + 2(0)7] <=0

Now, we put the value of f(1) in the equation, we get.

Rightarrow f(1)= (1)^ 7 [(1)^ 2 +1+2(1)^ 7 ] ⇒f(1) = = 1 1+1+2 1 4

hope it will help you