Question

find integrat x^2 +1/(x^2 +2)(x^2 +3)​

Answer 1

GIVEN :–

• A function –

$\\ \implies\bf\dfrac{x^2 +1}{(x^2 +2)(x^2 +3)}$

TO FIND :–

• Integration of function = ?

SOLUTION :–

• Let –

$\\ \implies\bf I = \int\dfrac{x^2 +1}{(x^2 +2)(x^2 +3)}dx$

• We should write this as –

$\\ \implies\bf I = \int \bigg[\dfrac{2}{(x^2 +3)} - \dfrac{1}{(x^2 +2)} \bigg]dx$

$\\ \implies\bf I =2\int\dfrac{1}{(x^2 +3)}dx -\int\dfrac{1}{(x^2 +2)}dx$

$\\ \implies\bf I =2I_1-I_2 \: \: \: \: \: \: \: \: - - - eq.(1)$

• Now Let's solve –

$\\ \implies\bf I_1 =\int\dfrac{1}{(x^2 +3)}dx$

• put x = t√3 –

$\\ \implies\bf dx = dt \sqrt{3}$

• So that –

$\\ \implies\bf I_1 =\int\dfrac{1}{ \{(t \sqrt{3})^2 +3 \}}(dt \sqrt{3})$

$\\ \implies\bf I_1 =\int\dfrac{1}{ \{3(t)^2 +3 \}}(dt \sqrt{3})$

$\\ \implies\bf I_1 = \dfrac{1}{ \sqrt{3} } \int\dfrac{1}{t^2 +1}dt$

$\\ \implies\bf I_1 = \dfrac{1}{ \sqrt{3} } \tan^{ - 1} (t)$

• Now replace 't' –

$\\ \implies\bf I_1 = \dfrac{1}{ \sqrt{3} } \tan^{ - 1} \bigg(\dfrac{x}{ \sqrt{3} } \bigg)$

• And –

$\\ \implies\bf I_2 =\int\dfrac{1}{(x^2 +2)}dx$

• put x = t√2 –

$\\ \implies\bf dx = dt \sqrt{2}$

• So that –

$\\ \implies\bf I_2 =\int\dfrac{1}{ \{(t \sqrt{2})^2 +2 \}}(dt \sqrt{2})$

$\\ \implies\bf I_2 =\int\dfrac{1}{ \{2(t)^2 +2\}}(dt \sqrt{2})$

$\\ \implies\bf I_2 = \dfrac{1}{ \sqrt{2} } \int\dfrac{1}{t^2 +1}dt$

$\\ \implies\bf I_2 = \dfrac{1}{ \sqrt{2} } \tan^{ - 1} (t)$

• Now replace 't' –

$\\ \implies\bf I_2 = \dfrac{1}{ \sqrt{2} } \tan^{ - 1} \bigg(\dfrac{x}{ \sqrt{2} } \bigg)$

• Put the values of $\tt I_1 \:\: and \:\: I_2$ in eq.(1) –

$\\ \implies \large{ \boxed{\bf I =\dfrac{2}{ \sqrt{3} } \tan^{ - 1} \bigg(\dfrac{x}{ \sqrt{3} } \bigg)- \dfrac{1}{ \sqrt{2} } \tan^{ - 1} \bigg(\dfrac{x}{ \sqrt{2} } \bigg) + c}}$