Evaluate- -dx-x2x4-13-4

The correct option is C ( 1 + 1/x^4 )^1/4 + c #160;I=∫ #160; dx / x^ 2 (x^ 4 +1)^ 3/4 Put #160; 1 / x ^ 4 =t⇒ 4 / x ^ 5 dx=dt # ...

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Standard XII

Mathematics

Special Integrals - 1

Evaluate: ∫...

Question

Evaluate: $\int \frac{d x}{x^{2} (x^{4} + 1)^{3 / 4}}$

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

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- {(1 - \frac{1}{x^{2}})}^{1 / 2} + c

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{(1 - \frac{1}{x^{2}})}^{1 / 2} + c

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

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Solution

The correct option is C $- {(1 + \frac{1}{x^{4}})}^{1 / 4} + c$
$I = \int \frac{d x}{x^{2} (x^{4} + 1)^{3 / 4}}$
Put $\frac{1}{x^{4}} = t \Rightarrow - \frac{4}{x^{5}} d x = d t$
$I = - \frac{1}{4} \int \frac{1}{{(t + 1)}^{3 / 4}} d t = - \sqrt[4]{t + 1}$
$= - \frac{\sqrt[4]{x^{4} + 1}}{x}$

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Similar questions

\int \frac{1}{x^{2} (x^{4} + 1)^{\frac{3}{4}}} d x

is equal to

\int e^{x} [\frac{x^{3} + x + 1}{(1 + x^{2})^{3 / 2}}] d x

is equal to

Q. Solve

\int \frac{d x}{x^{2} (x^{4} + 1)^{\frac{3}{4}}} = - {(1 + \frac{1}{x^{4}})}^{\frac{1}{m}}

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Q. The value of

\int \frac{(x^{4} - x)^{\frac{1}{4}}}{x^{5}} d x

$l i m x \to - 1 \frac{(1 + x) (1 - x^{2}) (1 + x^{3}) (1 - x^{4}) . . . . (1 - x^{4 n})}{{[(1 + x) (1 - x^{2}) (1 + x^{3}) (1 + x^{4}) . . . . . . . (1 - x^{2 n})]}^{2}} is equal to:$

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Evaluate: ∫dxx2(x4+1)3/4

The correct option is C −(1+1x4)1/4+cI=∫dxx2(x4+1)3/4Put 1x4=t⇒−4x5dx=dtI=−14∫1(t+1)3/4dt=−4√t+1=−4√x4+1x

Evaluate: $\int \frac{d x}{x^{2} (x^{4} + 1)^{3 / 4}}$

The correct option is C $- {(1 + \frac{1}{x^{4}})}^{1 / 4} + c$
$I = \int \frac{d x}{x^{2} (x^{4} + 1)^{3 / 4}}$
Put $\frac{1}{x^{4}} = t \Rightarrow - \frac{4}{x^{5}} d x = d t$
$I = - \frac{1}{4} \int \frac{1}{{(t + 1)}^{3 / 4}} d t = - \sqrt[4]{t + 1}$
$= - \frac{\sqrt[4]{x^{4} + 1}}{x}$