Evaluate:

Question:

Evaluate: $\int \frac{x^{3}}{x-2} d x$

Solution:

By doing long division of the given equation we get

Quotient $=x^{2}+2 x+4$

Remainder $=8$

$\therefore$ We can write the above equation as

$\Rightarrow x^{2}+2 x+4+\frac{8}{x-2}$

$\therefore$ The above equation becomes

$\Rightarrow \int \mathrm{x}^{2}+2 \mathrm{x}+4+\frac{8}{\mathrm{x}-2} \mathrm{dx}$

$\Rightarrow \int \mathrm{x}^{2} \mathrm{dx}+2 \int \mathrm{x} \mathrm{dx}+4 \int \mathrm{dx}+8 \int \frac{1}{\mathrm{x}-2} \mathrm{dx}$

We know $\int \mathrm{x} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}}}{\mathrm{n}+1} ; \int \frac{1}{\mathrm{x}} \mathrm{dx}=\ln \mathrm{x}$

$\Rightarrow \frac{x^{3}}{3}+2 \frac{x^{2}}{2}+4 x+8 \ln (x-2)+c$

$\Rightarrow \frac{x^{3}}{3}+x^{2}+4 x+8 \ln (x-2)+c \cdot($ Where $c$ is some arbitrary constant)

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