The primitive of frac{3x^4 - 1}{(x^4 + x + 1) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Divide the numerator and denominator by $x^4$: $I = \int \frac{3 - x^{-4}}{(1 + x^{-3} + x^{-4

Vedclass · Accepted Answer

$-\frac{x}{x^4 + x + 1} + c$

Vedclass · Answer

(B) Let $I = \int \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Divide the numerator and denominator by $x^4$: $I = \int \frac{3 - x^{-4}}{(1 + x^{-3} + x^{-4})^2} dx$. This does not simplify easily. Let us rewrite the integrand by dividing numerator and denominator by $x^2$: $I = \int \frac{3x^2 - x^{-2}}{(x^2 + 1 + x^{-1})^2} dx$. Let $u = x^2 + 1 + x^{-1}$. Then $du = (2x - x^{-2}) dx$. This is not matching. Let us try $u = \frac{x}{x^4 + x + 1}$. Using the quotient rule,$du = \frac{(x^4 + x + 1)(1) - x(4x^3 + 1)}{(x^4 + x + 1)^2} dx = \frac{x^4 + x + 1 - 4x^4 - x}{(x^4 + x + 1)^2} dx = \frac{1 - 3x^4}{(x^4 + x + 1)^2} dx$. Thus,$du = - \frac{3x^4 - 1}{(x^4 + x + 1)^2} dx$. Therefore,$I = - \int du = -u + c = -\frac{x}{x^4 + x + 1} + c$.

The primitive of $\frac{3x^4 - 1}{(x^4 + x + 1)^2}$ with respect to $x$ is:

Similar Questions

Let $g(x)$ be an antiderivative for $f(x)$. Then $\ln(1 + (g(x))^2)$ is an antiderivative for

$\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=A \cos x+B \log |f(x)|+c$ (where $c$ is a constant of integration). Then the values of $A, B$ and $f(x)$ are:

$\int \frac{3\cos x + 2\sin x}{4\sin x + 5\cos x} dx = A \{23x + 2\ln |4\sin x + 5\cos x|\} + c$,then $A$ and $f(x)$ are

If $\int \frac{2 x^2+5 x+9}{\sqrt{x^2+x+1}} d x=x \sqrt{x^2+x+1}+\alpha \sqrt{x^2+x+1}+\beta \log _{ e }\left| x +\frac{1}{2}+\sqrt{ x ^2+ x +1}\right|+ C$,where $C$ is the constant of integration,then $\alpha+2 \beta$ is equal to . . . . . .

$\int_0^{\pi /2} {\left( {\frac{\theta }{{\sin \theta }}} \right)^2 d\theta = } $

Vedclass Test Series

Exam Paper Generator

Online Exam Module