The integral {2 x³-1}{x^{4}+x} d x is equal to

Correct Answer is : log⁡∣x3+1x∣+c ≤ft|{x³+1}{x}|+clogxx3+1+c, (where c is a constant of integration)

{≤ft|x³+1|}{x²}+c₁ (where c is a constant of integration)

1/2 {≤ft(x³+1)²}{≤ft|x³|}+c, (where c is a constant of integration)

≤ft|{x³+1}{x}|+c, (where c is a constant of integration)

1/2 {≤ft|x³+1|}{x²}+c, (where c is a constant of integration)

Correct Answer is : log⁡∣x3+1x∣+c ≤ft|{x³+1}{x}|+clogxx3+1+c, (where c is a constant of integration)

Test Index

MHTCET 16 May 2024 Shift 1 Paper

Question : 147 of 150

Marks: +1, -0

The integral

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

is equal to

$\log \frac{\left|x^{3}+1\right|}{x^{2}}+c_{1}$ (where c is a constant of integration)
$\frac{1}{2} \log \frac{\left(x^{3}+1\right)^{2}}{\left|x^{3}\right|}+c$ , (where c is a constant of integration)
$\log \left|\frac{x^{3}+1}{x}\right|+c$ , (where c is a constant of integration)
$\frac{1}{2} \log \frac{\left|x^{3}+1\right|}{x^{2}}+c$ , (where c is a constant of integration)

Solution:

Concept:

Rewrite the integrand by dividing numerator and denominator by

x^2

to obtain the form

\frac{f'(x)}{f(x)}

, whose integral is

\log |f(x)|

Explanation:

Let

I = \int \frac{2x^3 - 1}{x^4 + x} dx

Divide numerator and denominator by

x^2

I = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} dx

Put

t = x^2 + \frac{1}{x}

, so

dt = \left(2x - \frac{1}{x^2}\right) dx

Therefore,

I = \int \frac{dt}{t} = \log |t| + c

Substituting back

t = x^2 + \frac{1}{x}

, we get:

I = \log \left|x^2 + \frac{1}{x}\right| + c = \log \left|\frac{x^3 + 1}{x}\right| + c

Answer:

Option C:

\log \left|\frac{x^3+1}{x}\right| + c

Go to Question:

Prev Question

Next Question