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Question

Evaluate $$\int { \cfrac { { x }^{ 3 }+{ 4x }^{ 2 }-7x+5 }{ x+2 } dx } $$


Solution

$$\int\dfrac{x^{3}+4x^{2}-7x+5}{x+2}dx$$

$$u=x+2, du=dx$$.

$$x=u-2$$
$$x^{2}=(u-2)^{2}$$=$$u^{2}+4-4u$$

$$x^{3}=(u-2)^{3}$$=$$u^{3}-8-6u^{2}+12u$$

$$x^{3}+4x^{2}-7x+5=u^{3}-8-6u^{2}+12u+4u^{2}+16-16u-7u+14+5$$

=$$u^{3}-2u^{2}-11u+27$$

=$$\int\dfrac{u^{3}-2u^{2}-11u+27}{u}du$$

=$$\int (u^{2}-2u-11+\dfrac{27}{u})du$$

=$$\dfrac{u^{3}}{3}-{u}^{2}-11u+27\ln u+C$$

=$$\dfrac{(x+2)^{3}}{3}-(x+2)^{2}-11(x+2)+27|\ln x+2\mid+C$$(Ans)

Mathematics

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