int frac{x^5+x}{x^8+1} dx = | vedclass

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

Question

(A) Let $I = \int \frac{x^5+x}{x^8+1} dx$. Divide numerator and denominator by $x^6$: $I = \int \frac{\frac{1}{x} + \frac{1}{x^5}}{x^2 + \frac{1}{x^6}

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} 	an ^{-1}\left(\frac{x^4-1}{\sqrt{2} x^2}ight)+c$

Vedclass · Answer

(A) Let $I = \int \frac{x^5+x}{x^8+1} dx$. Divide numerator and denominator by $x^6$: $I = \int \frac{\frac{1}{x} + \frac{1}{x^5}}{x^2 + \frac{1}{x^6}} dx$. This approach is complex. Alternative: Let $x^2 = t$, then $2x dx = dt$. $I = \frac{1}{2} \int \frac{t^2+1}{t^4+1} dt$. Divide numerator and denominator by $t^2$: $I = \frac{1}{2} \int \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} dt$. Let $t - \frac{1}{t} = u$, then $(1 + \frac{1}{t^2}) dt = du$. Also, $t^2 + \frac{1}{t^2} = (t - \frac{1}{t})^2 + 2 = u^2 + 2$. $I = \frac{1}{2} \int \frac{du}{u^2 + 2} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} 	an^{-1} \left(\frac{u}{\sqrt{2}}ight) + c$. Substituting $u = t - \frac{1}{t} = \frac{t^2-1}{t} = \frac{x^4-1}{x^2}$: $I = \frac{1}{2\sqrt{2}} 	an^{-1} \left(\frac{x^4-1}{\sqrt{2}x^2}ight) + c$.

$\int \frac{x^5+x}{x^8+1} dx =$

Similar Questions

If $\int \frac{a \cos x-2 \sin x}{b \sin x+5 \cos x} d x=\frac{7}{41} x+\frac{22}{41} \log |b \sin x+5 \cos x|+C, (a>0, b>0)$, then $\int \frac{d x}{b+a \cos x}=$

If $\int {\frac{{{x^4} + 1}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx = A \ln |x| + \frac{B}{{1 + {x^2}}} + c$,where $c$ is the constant of integration,then:

If $I_n = \int \frac{\sin nx}{\cos x} dx$,then $I_n =$

If $\tan \alpha = \frac{4}{3}$, then $\int \frac{1}{3 \cos x - 4 \sin x} dx = $

If $\int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}} = A \cdot \tan^{-1} \sqrt{f(x)} + \text{constant}$, then the ordered pair $(A, f(-1)) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module