Blog: `int (4x^3 + 3)/(x^4 + 3x) dx` Find the indefinite integral.

$\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}$

To solve, apply u-substitution method. So let:

$\displaystyle{u}={{x}}^{{4}}+{3}{x}$

Then, differentiate it.

$\displaystyle{d}{u}={\left({4}{{x}}^{{3}}+{3}\right)}{\left.{d}{x}\right.}$

Plug-in them to the integral.

$\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{x}}^{{4}}+{3}{x}}}\cdot{\left({4}{{x}}^{{3}}+{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{u}}{d}{u}$

Then, apply the integral formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle={\ln}{\left|{u}\right|}+{C}$

And, substitute back $\displaystyle{u}={{x}}^{{4}}+{3}{x}$ .

$\displaystyle={\ln}{\left|{{x}}^{{4}}+{3}{x}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}={\ln}{\left|{{x}}^{{4}}+{3}{x}\right|}+{C}$ .

$\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}$

To solve, apply u-substitution method. So let:

$\displaystyle{u}={{x}}^{{4}}+{3}{x}$

Then, differentiate it.

$\displaystyle{d}{u}={\left({4}{{x}}^{{3}}+{3}\right)}{\left.{d}{x}\right.}$

Plug-in them to the integral.

$\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{{{x}}^{{4}}+{3}{x}}}\cdot{\left({4}{{x}}^{{3}}+{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{1}}{{u}}{d}{u}$

Then, apply the integral formula $\displaystyle\int\frac{{1}}{{x}}{\left.{d}{x}\right.}={\ln}{\left|{x}\right|}+{C}$ .

$\displaystyle={\ln}{\left|{u}\right|}+{C}$

And, substitute back $\displaystyle{u}={{x}}^{{4}}+{3}{x}$ .

$\displaystyle={\ln}{\left|{{x}}^{{4}}+{3}{x}\right|}+{C}$

Therefore, $\displaystyle\int\frac{{{4}{{x}}^{{3}}+{3}}}{{{{x}}^{{4}}+{3}{x}}}{\left.{d}{x}\right.}={\ln}{\left|{{x}}^{{4}}+{3}{x}\right|}+{C}$ .

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