Question

Find the following integral : $\int \frac{x^3+x^2 +x-1}{x-1} \, dx$

Answer 1

Answer:

Integral of the given function is $\frac{x^3}{3}+x^2+3x+2log(x-1)+c$

Step-by-step explanation:

Concept:

I have applied decomposition method to solve this problem.

In decomposition method, the given non-integrable function is decomposed into integrable function by using algebraic identities, trigonometric identities, etc.

Formula used:

$a^3-b^3=(a-b)(a^2+ab+b^2)\\\\a^2-b^2=(a-b)(a+b)$

Now,

$\int{\frac{x^3+x^2+x_1}{x-1}}\:dx$

$=\int{\frac{(x^3-1)+(x^2-1)+(x-1)+2}{x-1}}\:dx\\\\=\int{[\frac{x^3-1^3}{x-1}+\frac{x^2-1^2}{x-1}+\frac{x-1}{x-1}+\frac{2}{x-1}]\:dx$

$=\int{[\frac{(x-1)(x^2+x+1))}{x-1}+\frac{(x-1)(x+1)}{x-1}+1+\frac{2}{x-1}]\:dx$

$=\int{[x^2+x+1+x+1+1+\frac{2}{x-1}]\:dx$

$=\int{[x^2+2x+3+\frac{2}{x-1}]\:dx$

$=\int{x^2}\:dx+2\int{x}\:dx+3\int\:dx+2\int{\frac{1}{x-1}}\:dx\\\\=\frac{x^3}{3}+2\frac{x^2}{2}+3x+2log(x-1)+c\\\\=\frac{x^3}{3}+x^2+3x+2log(x-1)+c$