Question

Q-1) Find :-

${ \bf{∫ \frac{2x}{(x {}^{2} + 1)(x {}^{2} + 2) {}^{2} }dx}}$
​

Answer 1

☯ Explanation,

$\displaystyle \tt \int \dfrac{2x}{(x {}^{2} + 1)(x {}^{2} + 2) {}^{2} } dx \\ \\ \\ \leadsto \displaystyle2 \tt \int \dfrac{x}{( {x}^{2} + 1)(x {}^{2} + 2) {}^{2} } dx \\ \\ \\ \tt \: Now, \: take \: the \: partial \: fraction. \\ \\ \\ \tt \frac{x}{x {}^{2} - 1 } - \frac{x}{x {}^{2} + 2 } - \dfrac{x}{(x {}^{2} + 2) {}^{2} } \\ \\ \\ \tt \leadsto \displaystyle2 \tt \int \frac{x}{x {}^{2} - 1 } - \frac{x}{x {}^{2} + 2 } - \dfrac{x}{(x {}^{2} + 2) {}^{2} } dx \\ \\ \\ \tt Applying \: sum \: rule, \: we \: get; \\ \\ \\ \leadsto\displaystyle2 \tt \int \frac{x}{x {}^{2} - 1 } dx - \int \frac{x}{x {}^{2} + 2 } dx - \int \dfrac{x}{(x {}^{2} + 2) {}^{2} } dx \\ \\ \\ \leadsto \tt \: 2( \dfrac{1}{2}ln|x {}^{2} + 1| - \dfrac{1}{2}ln |x {}^{2} - 2| - \bigg( - \dfrac{ 1 }{2(x {}^{2} - 2) } \bigg) \\ \\ \\ \leadsto \tt \: ln|x {}^{2} + 1| - ln|x {}^{2} + 2| + \frac{1}{x {}^{2} + 2 } \\ \\ \\ \leadsto \underline{ \boxed{ \tt ln|x {}^{2} + 1| - ln|x {}^{2} + 2| + \frac{1}{x {}^{2} + 2 } + c}}$

Answer 2

Solution

$\implies\tt \int \dfrac{2x}{( {x}^{2} + 1)( {x}^{2} + 2)^{2} } dx$

Now Using Substitution Method

$\tt \implies \: Let ,\: \: \: {x}^{2} + 2 = t$

$\implies \tt \: \dfrac{d( {x}^{2} + 2) }{dx} = \dfrac{dt}{dx}$

$\tt \implies \: 2x = \dfrac{dt}{dx}$

$\tt \implies(2x)dx = dt$

Now Put the value

$\implies\tt \int \dfrac{2x}{( {x}^{2} + 1)( {x}^{2} + 2)^{2} } dx$

To eliminate x² + 1 , We can write as

$\tt \implies \: {x}^{2} + 2 = t$

$\tt \implies \: {x}^{2} + 1 + 1 = t$

$\implies \tt {x}^{2} + 1 = t - 1$

Now we can write as

$\tt \implies \: \int \dfrac{dt}{(t - 1) {t}^{2} }$

$\tt \implies \: \int \dfrac{t - (t - 1)}{(t - 1) {t}^{2} } dt$

$\tt \implies \: \int \dfrac{t}{(t - 1) {t}^{2} } dt - \int \dfrac{(t - 1)}{(t - 1) {t}^{2} } dt$

$\tt \implies \int \dfrac{dt}{(t - 1)t} - \int \dfrac{1}{ {t}^{2} } dt$

$\tt \implies \int \dfrac{t - (t - 1)}{(t - 1)t} dt - \int {t}^{ - 2} dt$

$\tt \implies \: \int \frac{t}{(t - 1)t} dt - \int \frac{(t - 1)}{t(t - 1)} dt - \bigg( \dfrac{t {}^{ - 2 + 1} }{ - 2 + 1} \bigg) + c$

$\tt \implies \int \dfrac{dt}{(t - 1)} - \int \dfrac{1}{t} dt + \dfrac{1}{t} + c$

$\tt \implies log |t - 1| - log |t| + \dfrac{1}{t} + c$

Now put the value of t = x² + 2

$\tt \implies \: log | {x}^{2} + 2 - 1| - log | {x}^{2} + 2 | + \dfrac{1}{ {x}^{2} + 2 } + c$

$\tt \implies \: log | {x}^{2} + 1| - log | {x}^{2} + 2 | + \dfrac{1}{ {x}^{2} + 2 } + c$

Answer

$\tt \implies \: log | {x}^{2} + 1| - log | {x}^{2} + 2 | + \dfrac{1}{ {x}^{2} + 2 } + c$