The value of {d x}{x²≤ft(x^{4}+1)³/4}} is

Correct Answer is : −(x4+1x4)14+c-≤ft({x^{4}+1}{x^{4}})¹/4}+c−(x4x4+1)41+c, where ccc is constant of integration.

≤ft({-x^{4}+1}{x^{4}})¹/4}+c, where c is constant of integration.

≤ft(x^{4}+1)¹/4}+c, where c is constant of integration.

-≤ft(x^{4}+1)¹/4}+c, where c is constant of integration.

-≤ft({x^{4}+1}{x^{4}})¹/4}+c, where c is constant of integration.

Correct Answer is : −(x4+1x4)14+c-≤ft({x^{4}+1}{x^{4}})¹/4}+c−(x4x4+1)41+c, where ccc is constant of integration.

Test Index

MHTCET 10 May 2024 Shift 2 Paper

Question : 144 of 150

Marks: +1, -0

The value of

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

$\left(\frac{-x^{4}+1}{x^{4}}\right)^{\frac{1}{4}}+c$ , where $c$ is constant of integration.
$\left(x^{4}+1\right)^{\frac{1}{4}}+c$ , where c is constant of integration.
$-\left(x^{4}+1\right)^{\frac{1}{4}}+c$ , where $c$ is constant of integration.
$-\left(\frac{x^{4}+1}{x^{4}}\right)^{\frac{1}{4}}+c$ , where $c$ is constant of integration.

Solution:

Concept:

Rewrite the integrand by taking

x^4

common from

(x^4+1)

and then use a suitable substitution.

Explanation:

Let

I = \int \frac{dx}{x^2(x^4+1)^{3/4}}

Write

(x^4+1)^{3/4} = \left[x^4\left(1+\frac{1}{x^4}\right)\right]^{3/4} = x^3\left(1+\frac{1}{x^4}\right)^{3/4}

So,

I = \int \frac{dx}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = \int x^{-5}\left(1+\frac{1}{x^4}\right)^{-3/4}dx

Put

t = 1 + \frac{1}{x^4}

Then,

dt = -\frac{4}{x^5}dx

, hence

x^{-5}dx = -\frac{1}{4}dt

Therefore,

I = -\frac{1}{4}\int t^{-3/4}dt = -t^{1/4}+c

Substitute back

t = 1+\frac{1}{x^4}

Thus,

I = -\left(\frac{x^4+1}{x^4}\right)^{1/4}+c

Answer:

Option D:

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

, where

c

is the constant of integration.

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