Question

homework 2: 1.5 substitution
score: 10/35 answered: 4/9
question 5
consider the indefinite integral $int \frac{8x^{7}}{sqrt{1 + 9x^{8}}}dx$.
a) this can be transformed using the substitution
$u = square$
which gives $du = square$ (dont forget the differential $dx$ or $du$.)
c) performing the substitution in terms of $u$ gives the integral
$int square$
d) evaluate the integral and simplify. your answer should be in terms of $x$, not $u$.
$square + c$

Explanation:

Step1: Choose substitution variable

Let $u = 1 + 9x^8$

Step2: Compute derivative of u

$du = 72x^7 dx$

Step3: Rewrite integral in terms of u

First, solve for $8x^7 dx$: $\frac{1}{9}du = 8x^7 dx$. Substitute into integral:
$\int \frac{1}{\sqrt{u}} \cdot \frac{1}{9} du = \frac{1}{9} \int u^{-\frac{1}{2}} du$

Step4: Evaluate the u-integral

Use power rule: $\int u^n du = \frac{u^{n+1}}{n+1} + C$
$\frac{1}{9} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = \frac{2}{9}u^{\frac{1}{2}} + C$

Step5: Substitute back to x

Replace $u$ with $1 + 9x^8$:
$\frac{2}{9}\sqrt{1 + 9x^8} + C$

Answer:

a) $u = 1 + 9x^8$
$du = 72x^7 dx$
c) $\frac{1}{9} \int u^{-\frac{1}{2}} du$
d) $\frac{2}{9}\sqrt{1 + 9x^8}$