Question

chapter - integration ​

Answer 1

Given Integrand,

$f(x) = \displaystyle \int \dfrac{ {2x}^{3} - 1}{x + {x}^{4} } dx$

Dividing with x^2 in both numerator and denominator,

$\longrightarrow \: f(x) = \displaystyle \int \dfrac{ {2x}^{} - \dfrac{1}{ {x}^{2} } }{ \dfrac{1}{x} + {x}^{2} } dx$

Let t = x^2 + 1/x.

Differentiating w.r.t x on both sides, we get :

$dx = \dfrac{dt}{2x - \dfrac{1}{ {x}^{2} } }$

Now,

$\longrightarrow \: f(x) = \displaystyle \int \dfrac{dt}{t} \\ \\ \longrightarrow \: f(x) = ln(t) + c \\ \\ \longrightarrow \: f(x) = ln \bigg( {x}^{2} + \dfrac{1}{x} \bigg) + c$

ALITER,

We know that,

$\displaystyle \int \dfrac{f'(x)}{f(x)} = ln(f(x)) + C$

Thus,

$\longrightarrow f(x) = ln \bigg( {x}^{2} + \dfrac{1}{x} \bigg) + c$

Answer 2

$\int \dfrac{2x^3-1}{x+x^4}dx$

$\displaystyle=\int \frac{2u-1}{3u^{\frac{2}{3}}\left(\sqrt[3]{u}+u^{\frac{4}{3}}\right)}du$

$\displaystyle =\frac{1}{3}\cdot \int \frac{2u-1}{u^{\frac{2}{3}}\left(\sqrt[3]{u}+u^{\frac{4}{3}}\right)}du$

$\displaystyle=\frac{1}{3}\cdot \int \frac{2u-1}{u+u^2}du$

$\displaystyle=\frac{1}{3}\left(\int \frac{2u}{u+u^2}du-\int \frac{1}{u+u^2}du\right)$

$=\dfrac{1}{3}\left(2\ln \left|1+u\right|-\left(\ln \left|u\right|-\ln \left|u+1\right|\right)\right)$

$=\dfrac{1}{3}\left(2\ln \left|1+x^3\right|-\left(\ln \left|x^3\right|-\ln \left|x^3+1\right|\right)\right)$

$=\dfrac{1}{3}\left(3\ln \left|x^3+1\right|-\ln \left|x^3\right|\right)$

$=\dfrac{1}{3}\left(3\ln \left|x^3+1\right|-\ln \left|x^3\right|\right)+C$