Integration using Substitution
Integration using Substitution
Calculation of \(\displaystyle \int \frac{5x^3+3x-1}{(x^3+3x+1)^3}dx\)
- Tolaso J Kos
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Re: Integration using Substitution
Regroupping terms we have that our integral can be re-written as follows:
$$\begin{aligned}
\int \frac{5x^3+3x-1}{\left ( x^3+3x+1 \right )^3}\,dx &=\int \frac{-x^3-3x-1+6x^3+6x}{\left ( x^3+3x+1 \right )^3}\,dx \\
&= \int \frac{-x^3-3x-1+2x\left ( 3x^2+3 \right )}{\left ( x^3+x+1 \right )^3}\,dx\\
&= \int \frac{-\left ( x^3+3x+1 \right )^2+2x\left ( x^3+3x+1 \right )\left ( 3x^2+3x \right )}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \frac{-(x)'\left ( x^3+3x+1 \right )^2+x\left[ \left ( x^3+3x+1 \right )^2 \right]'}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \left [ -\frac{x}{\left ( x^3+3x+1 \right )^2} \right ]' \,dx = -\frac{x}{\left ( x^3+3x+1 \right )^2}+c, \; \; c \in \mathbb{R}
\end{aligned}$$
By the way I would like to see a sub solution... because this is very technical and it is based on observation.
$$\begin{aligned}
\int \frac{5x^3+3x-1}{\left ( x^3+3x+1 \right )^3}\,dx &=\int \frac{-x^3-3x-1+6x^3+6x}{\left ( x^3+3x+1 \right )^3}\,dx \\
&= \int \frac{-x^3-3x-1+2x\left ( 3x^2+3 \right )}{\left ( x^3+x+1 \right )^3}\,dx\\
&= \int \frac{-\left ( x^3+3x+1 \right )^2+2x\left ( x^3+3x+1 \right )\left ( 3x^2+3x \right )}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \frac{-(x)'\left ( x^3+3x+1 \right )^2+x\left[ \left ( x^3+3x+1 \right )^2 \right]'}{\left ( x^3+3x+1 \right )^4}\,dx\\
&= \int \left [ -\frac{x}{\left ( x^3+3x+1 \right )^2} \right ]' \,dx = -\frac{x}{\left ( x^3+3x+1 \right )^2}+c, \; \; c \in \mathbb{R}
\end{aligned}$$
By the way I would like to see a sub solution... because this is very technical and it is based on observation.
Imagination is much more important than knowledge.
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