- dxx -13 x -251 - 4 is equal toA-34 x-1 x-41 - 4-CB- 43 x-1 x-41 - 4-CC- 35 x-1 x-41 - 4-CD53 x-1 x-41 - 4-C

This integral can also be written as nbsp; I=∫dxx 13x+251/4⇒I=∫dxx 1x+23/4(x+2)2 Now let, x 1x+2=t⇒dt=3dx(x+2)2 Thus our integral becomes, I=13∫1t3/4dt nbsp ...

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Byju's Answer

Standard XII

Mathematics

Integration of Irrational Algebraic Fractions - 2

∫ dxx -13 x +...

Question

$\int \frac{dx}{{({(x - 1)}^{3} {(x + 2)}^{5}]}^{1 / 4}}$ is equal to

\frac{3}{4} {(\frac{x - 1}{x + 4})}^{1 / 4} + C

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\frac{4}{3} {(\frac{x - 1}{x + 4})}^{1 / 4} + C

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\frac{3}{5} {(\frac{x - 1}{x + 4})}^{1 / 4} + C

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\frac{5}{3} {(\frac{x - 1}{x + 4})}^{1 / 4} + C

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Solution

The correct option is B
$\frac{4}{3} {(\frac{x - 1}{x + 4})}^{1 / 4} + C$
This integral can also be written as
$I = \int \frac{dx}{{({(x - 1)}^{3} {(x + 2)}^{5}]}^{1 / 4}} \Rightarrow I = \int \frac{dx}{{(\frac{x - 1}{x + 2})}^{3 / 4} (x + 2)^{2}}$
Now let,
$(\frac{x - 1}{x + 2}) = t \Rightarrow dt = \frac{3 dx}{(x + 2)^{2}}$
Thus our integral becomes,
$I = \frac{1}{3} \int \frac{1}{t^{3 / 4}} dt = \frac{1}{3} (\frac{t^{1 / 4}}{1 / 4}) + C = \frac{4}{3} t^{1 / 4} + C = \frac{4}{3} {(\frac{x - 1}{x + 2})}^{1 / 4} + C$
Thus, Option b. is correct.

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∫dx((x−1)3(x+2)5]1/4 is equal to

The correct option is B 43(x−1x+4)1/4+CThis integral can also be written as I=∫dx((x−1)3(x+2)5]1/4⇒I=∫dx(x−1x+2)3/4(x+2)2 Now let, (x−1x+2)=t⇒dt=3dx(x+2)2 Thus our integral becomes, I=13∫1t3/4dt =13(t1/41/4)+C =43t1/4+C =43(x−1x+2)1/4+C Thus, Option b. is correct.

$\int \frac{dx}{{({(x - 1)}^{3} {(x + 2)}^{5}]}^{1 / 4}}$ is equal to