If [f( x ) = 5x^8 + 7x^6 ( x^2 + 1 + 2x^7 )^2 ,dx,,( x 0 ),] [f( 0 ) = 0,] then the value …

Question

If   $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$

$$f\left( 0 \right) = 0,$$    then the value of $f(1)$ is :

Accepted Answer

$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$

$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$

$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$

Let ${x^{ - 5}} + {x^{ - 7}} + 2 = t$

$$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$$

$$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx =  - dt$$

$$f(x) = \int {{{ - dt} \over {{t^2}}}}  = {1 \over t} + c$$

$$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$$

$$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$$

$$f\left( 0 \right) = 0$$

$$ \therefore $$ $c = 0$

$$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$$

$$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$$

Solutions · 1