Find: x^ 2 (x^ 2 +1 ) (x^ 2 +4 ) d x - Vidyadip

Question

Find: $\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$

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Answer

Consider $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$ and put x² = y
Then, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}$
Write, $\frac{y}{(y+1)(y+4)}=\frac{A}{y+1}+\frac{B}{y+4}$
So that, y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides,
we get A + B = 1 and 4A + B = 0, which give
$A=-\frac{1}{3}$ and $B=\frac{4}{3}$
Thus, $\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3\left(x^{2}+1\right)}+\frac{4}{3\left(x^{2}+4\right)}$
= $-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$
= $-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1} \frac{x}{2}+C$

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