If m is a non-zero number and -x5 m -1-2 - x 4 m -1- x 2 m - x m -13 dx - f x - c- then f x isA- x5 m-2 mx2 m-xm-12B- x4n -2 mx2 m-xm-12C- 2 -xm-x-x2 m-xm-12D- x5 m-x4 m-2 mx2 m-xm-12

The correct option is B x^4m/2m(x^2m + x^m + 1)^2We have, ∫x^5m 1 + 2.x^4m 1/(x^2m + x^m + 1)^3dx =∫x^ m 1 + 2x^ 2m 1/(1 + 1/x^m + 1/x^2m)^3dx Let 1 + 1/x^m ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If m is a non...

Question

If $m$ is a non-zero number and $\int \frac{x^{5 m - 1} + 2. x^{4 m - 1}}{(x^{2 m} + x^{m} + 1)^{3}} d x = f (x) + c,$ then $f (x)$ is

\frac{x^{5 m}}{2 m (x^{2 m} + x^{m} + 1)^{2}}

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\frac{x^{4 m}}{2 m (x^{2 m} + x^{m} + 1)^{2}}

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\frac{2 m (x^{5 m} + x^{4 m})}{(x^{2 m} + x^{m} + 1)^{2}}

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\frac{(x^{5 m} - x^{4 m})}{2 m (x^{2 m} + x^{m} + 1)^{2}}

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Solution

The correct option is B $\frac{x^{4 m}}{2 m (x^{2 m} + x^{m} + 1)^{2}}$
We have,
$\int \frac{x^{5 m - 1} + 2. x^{4 m - 1}}{(x^{2 m} + x^{m} + 1)^{3}} d x$
$= \int \frac{x^{- m - 1} + 2 x^{- 2 m - 1}}{(1 + \frac{1}{x^{m}} + \frac{1}{x^{2 m}})^{3}} d x$
$L e t 1 + \frac{1}{x^{m}} + \frac{1}{x^{2 m}} = t$
$I = - \frac{1}{m} \int \frac{d t}{t^{3}} = - \frac{1}{m} (\frac{t^{- 2}}{- 2}) + c$
$= \frac{x^{4 m}}{2 m (x^{2 m + x^{m} + 1})^{2}} + c$

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Similar questions

Q. If

m

is a non-zero number and

\int \frac{x^{5 m - 1} + 2. x^{4 m - 1}}{(x^{2 m} + x^{m} + 1)^{3}} d x = f (x) + c,

then

f (x)

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$x^{m} + x^{m} = x^{2 m},$ where $x$ is a non-zero rational number and $m$ is a positive integer.

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$x^{m} \times x^{n} = x^{m + n},$ where $x$ is a non-zero rational number and $m, n$ are positive integers.

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If m is a non-zero number and ∫x5m−1+2.x4m−1(x2m+xm+1)3dx=f(x)+c, then f(x) is

The correct option is B x4m2m(x2m+xm+1)2We have, ∫x5m−1+2.x4m−1(x2m+xm+1)3dx =∫x−m−1+2x−2m−1(1+1xm+1x2m)3dx Let 1+1xm+1x2m=t I=−1m∫dtt3=−1m(t−2−2)+c =x4m2m(x2m+xm+1)2+c

If $m$ is a non-zero number and $\int \frac{x^{5 m - 1} + 2. x^{4 m - 1}}{(x^{2 m} + x^{m} + 1)^{3}} d x = f (x) + c,$ then $f (x)$ is