Question

assignment submission for this assignment, you submit answers by question parts. the number of assignment scoring your best submission for each question part is used for your score. 7. 0 / 2.08 points evaluate the integral. $int_{1}^{2}\frac{x^{3}+3x}{x^{4}+6x^{2}+2}dx$

Explanation:

Step1: Use substitution

Let $u = x^{4}+6x^{2}+2$. Then $du=(4x^{3}+12x)dx = 4(x^{3}+3x)dx$, so $(x^{3}+3x)dx=\frac{1}{4}du$.
When $x = 1$, $u=1 + 6+2=9$. When $x = 2$, $u=16 + 24+2=42$.

Step2: Rewrite the integral

The integral $\int_{1}^{2}\frac{x^{3}+3x}{x^{4}+6x^{2}+2}dx$ becomes $\frac{1}{4}\int_{9}^{42}\frac{du}{u}$.

Step3: Integrate

We know that $\int\frac{du}{u}=\ln|u|+C$. So $\frac{1}{4}\int_{9}^{42}\frac{du}{u}=\frac{1}{4}[\ln(u)]_{9}^{42}$.

Step4: Evaluate the definite - integral

$\frac{1}{4}(\ln(42)-\ln(9))=\frac{1}{4}\ln(\frac{42}{9})=\frac{1}{4}\ln(\frac{14}{3})$.

Answer:

$\frac{1}{4}\ln(\frac{14}{3})$