Finding the indefinite integral $\int \frac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$

Question

I'm not too familiar with how to solve this. Could anyone present a step by step guide on how to get the answer?

$$\int \dfrac{3x+2}{(6x^2+8x)^7}\,\mathrm dx$$

Answer 1

Let $u = 6x^{2} + 8x$. Then $du = (12x + 8) dx \rightarrow du = 4(3x + 2) dx$. This is equivalent to $\frac{du}{4} = (3x + 2) dx$

Make the substitution for the integral, so we have:

$$\int \frac{((3x + 2) dx)}{u^{7}}$$ $$= \int \frac{1/4 du}{u^7}$$ $$= \frac{1}{4} \int \frac{du}{u^7}$$ $$= \frac{1}{4} \int u^{-7} du$$

Thus, by power rule, we have:

$$\frac{1}{4} \frac{u^{-7 + 1}}{-7 + 1} + c$$ $$= \frac{1}{4} \frac{u^{-6}}{-6} + c$$ $$= \frac{-1}{24} u^{-6} + c$$ $$= \frac{-1}{24} (6x^{2} + 8x)^{-6} + c \text{ where c is an arbitrary constant}$$

In terms of positive exponent, we obtain:

$$\frac{-1}{24(6x^{2} + 8x)^{6}} + c$$

Answer 2

Put $6x^2+8x=u\implies (12x+8)dx=du \implies 4(3x+2)dx=du$

So, $$\int \dfrac{3x+2}{(6x^2+8x)^7}\,\mathrm dx=\int\frac{du}{4u^7}=\frac14\int u^{-7}du=\frac14\cdot \frac{u^{-7+1}}{-7+1}+C$$ where $C$ is an arbitrary constant of indefinite integration .