MHB Integral Evluation: $$\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$$

[juantheron]

Evaluation of $$\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$$$\bf{My\;Try::}$ Let $\displaystyle f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now Diff. both side w. r to $x\;,$ We Get$\displaystyle \Rightarrow f'(x) = \left\{\frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot (ax+b)}{(x^3+3x+1)^4}\right\} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$So $\displaystyle \frac{(x^3+3x+1)\cdot a-6\cdot (x^2+1)\cdot (ax+b)}{(x^3+3x+1)^3} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}.$So $\displaystyle \frac{-5ax^3+6bx^2-3ax+(a-6b)}{(x^3+3x+1)^3} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$.So We Get $a = -1$ and $b = 0.$ So We Get $\displaystyle \frac{d}{dx}\left[f(x)\right] = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$So $\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx = \int\frac{d}{dx}\left[f(x)\right]dx = f(x) = -\frac{x}{(x^3+3x+1)^2}+\mathcal{C}$But How we can Solve the above integral Directly.

Means Adding and subtracting in Numerator and taking something common

and Then Using Substitution Method , Thanks.

 

[Prove It]

Evaluation of $$\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx$$$\bf{My\;Try::}$ Let $\displaystyle f(x) = \frac{ax+b}{(x^3+3x+1)^2}.$ Now Diff. both side w. r to $x\;,$ We Get$\displaystyle \Rightarrow f'(x) = \left\{\frac{(x^3+3x+1)^2\cdot a-2\cdot (x^3+3x+1)\cdot (3x^2+3)\cdot (ax+b)}{(x^3+3x+1)^4}\right\} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$So $\displaystyle \frac{(x^3+3x+1)\cdot a-6\cdot (x^2+1)\cdot (ax+b)}{(x^3+3x+1)^3} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}.$So $\displaystyle \frac{-5ax^3+6bx^2-3ax+(a-6b)}{(x^3+3x+1)^3} = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$.So We Get $a = -1$ and $b = 0.$ So We Get $\displaystyle \frac{d}{dx}\left[f(x)\right] = \frac{5x^3+3x-1}{(x^3+3x+1)^3}$So $\displaystyle \int\frac{5x^3+3x-1}{(x^3+3x+1)^3}dx = \int\frac{d}{dx}\left[f(x)\right]dx = f(x) = -\frac{x}{(x^3+3x+1)^2}+\mathcal{C}$But How we can Solve the above integral Directly.

Means Adding and subtracting in Numerator and taking something common

and Then Using Substitution Method , Thanks.

$\displaystyle \begin{align*} \int{ \frac{5x^3 + 3x - 1}{ \left( x^3 + 3x + 1 \right) ^3} \,\mathrm{d}x} &= \int{ \frac{\left( x^3 + 3x + 1 \right) \left( 5x^3 + 3x - 1 \right) }{ \left( x^3 + 3x + 1 \right) ^4 }\,\mathrm{d}x} \\ &= \int{ \frac{5x^6 + 18x^4 + 4x^3 + 9x^2 - 1}{ \left[ \left( x^3 + 3x + 1 \right) ^2 \right] ^2} \,\mathrm{d}x} \end{align*}$

Now because we have a square on the bottom, we MAY be able to make it look like a quotient rule expansion, recall $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \, \left( \frac{u}{v} \right) = \frac{\frac{\mathrm{d}u}{\mathrm{d}x}\,v - u\,\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \end{align*}$, so we require $\displaystyle \begin{align*} v = \left( x^3 + 3x + 1 \right) ^2 \end{align*}$ and thus

$\displaystyle \begin{align*} \left( x^3 + 3x + 1 \right) ^2 \, \frac{\mathrm{d}u}{\mathrm{d}x} + 2\,\left( 3x^2 + 3 \right) \left( x^3 + 3x + 1 \right) \, u &= 5x^6 + 18x^4 + 4x^3 + 9x^2 - 1 \\ \frac{\mathrm{d}}{\mathrm{d}x} \, \left[ \left( x^3 + 3x + 1 \right) ^2 \, u \right] &= 5x^6 + 18x^4 + 4x^3 + 9x^2 - 1 \\ \left( x^3 + 3x + 1 \right) ^2 \, u &= \int{ 5x^6 + 18x^4 + 4x^3 + 9x^2 - 1 \,\mathrm{d}x } \\ \left( x^3 + 3x + 1 \right) ^2 \, u &= \frac{5x^7}{7} + \frac{18x^5}{5} + x^4 + 3x^3 - x + C \end{align*}$

Once you have found u, you should be able to write your original integrand as the derivative of a product. Then the answer will be obvious.