If -8x43 - 13-x38-x13 - x5 - 14dx - 1-3- xa-xb - xc - 13-C where a- b- c - - a-b-c and where C is a constant of integration- then

I=∫8/x^9 + 13/x^14/(1 + 1/x^8 + 1/x^13)^4dx Put, 1 + 1/x^8 + 1/x^13 = t ⇒ nbsp; ( 8/x^9 13/x^14)dx = dt ⇒ I=∫ dt/t^4 = 1/3t^3 + C = 1/3x^39/(x^13 + x^5 + 1 ...

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Byju's Answer

Standard XI

Mathematics

Integration by Substitution

If ∫8x43 + 13...

Question

If $\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C$ where $a, b, c ϵ N, (a > b > c$ and where C is a constant of integration), then

a + b = 39

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a + b = 52

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a + b + c = 57

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b - c = 8

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Solution

The correct option is D $b - c = 8$
$I = \int \frac{\frac{8}{x^{9}} + \frac{13}{x^{14}}}{(1 + \frac{1}{x^{8}} + \frac{1}{x^{13}})^{4}} d x Put, 1 + \frac{1}{x^{8}} + \frac{1}{x^{13}} = t \Rightarrow (\frac{- 8}{x^{9}} - \frac{13}{x^{14}}) d x = d t \Rightarrow I = \int \frac{- d t}{t^{4}} = \frac{1}{3 t^{3}} + C = \frac{1}{3} \frac{x^{39}}{(x^{13} + x^{5} + 1)^{3}} + C a = 39, b = 13, c = 5$

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Similar questions

Q. If

\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C

where

a, b, c ϵ N, (a > b > c

and where C is a constant of integration), then

Q. If

\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C

where

a, b, c ϵ N, (a > b > c

and where C is a constant of integration), then

{(\begin{matrix} \frac{x^{a}}{x^{b}} \end{matrix})}^{a + b} \times {(\begin{matrix} \frac{x^{b}}{x^{c}} \end{matrix})}^{b + c} \times {(\begin{matrix} \frac{x^{c}}{x^{a}} \end{matrix})}^{c + a}

= ____.

\frac{1}{1 + x^{(b - a)} + x^{(c - a)}} + \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} + \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ?

{(\begin{matrix} \frac{x^{b}}{x^{c}} \end{matrix})}^{(b + c - a)} . {(\begin{matrix} \frac{x^{c}}{x^{a}} \end{matrix})}^{(c + a - b)} . x {(\begin{matrix} \frac{x^{a}}{x^{b}} \end{matrix})}^{(a + b - c)} = ?

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If ∫8x43+13.x38(x13+x5+1)4dx=13.xa(xb+xc+1)3+C where a,b,c ϵ N,(a>b>c and where C is a constant of integration), then

The correct option is D b−c=8I=∫8x9+13x14(1+1x8+1x13)4dxPut,1+1x8+1x13=t⇒(−8x9−13x14)dx=dt⇒I=∫−dtt4=13t3+C=13x39(x13+x5+1)3+Ca=39,b=13,c=5

If $\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C$ where $a, b, c ϵ N, (a > b > c$ and where C is a constant of integration), then