Help on indefinite integration needed!

**lightbird** · June 2nd 2008, 03:08 AM

Well, there will be several ones. It were really helpful if someone could also explain some details;

1) I (x^3-x^2-3x+1)/(x^2+x-2)dx

2) I sin4x*sin5x dx

3) I e^cosx sinx dx

4) I x artctgx dx

5) I (1/x^2)*e^1/x dx

**Soroban** · June 2nd 2008, 04:00 AM

Hello, lightbird!

Here are a few of them . . .

$1)\;\;\int \frac{x^3-x^2-3x+1}{x^2+x-2}\,dx$

Long division: . $\frac{x^3-x^2-3x+1}{x^2+x-2} \;=\;x - 2 + \frac{x-3}{x^2+x-2}$

$\text{Then we have: }\;\int(x + 2)\,dx + \underbrace{\int\frac{x-3}{(x-1)(x+2)}\,dx}_{\text{Partial fractions}}$

$3)\;\;\int e^{\cos x}\sin x\,dx$

Let $u \:=\:\cos x\quad\Rightarrow\quad du \:=\:-\sin x\,dx\quad\Rightarrow\quad \sin x\,dx \:=\:-du$

Substitute: . $\int e^u(-du) \;=\;-\int e^u\,du$

$5)\;\;\int \frac{1}{x^2}\cdot e^{\frac{1}{x}}\,dx$

Let $u \:=\: \frac{1}{x} \:=\: x^{-1}\quad\Rightarrow\quad du \:=\:-x^{-2}dx\quad\Rightarrow\quad \frac{1}{x^2}dx \:=\:-du$

Substitute: . $\int e^u(-du) \;=\;-\int e^u\,du$

**mr fantastic** · June 2nd 2008, 04:06 AM

Originally Posted by lightbird

Well, there will be several ones. It were really helpful if someone could also explain some details;

1) I (x^3-x^2-3x+1)/(x^2+x-2)dx

2) I sin4x*sin5x dx

3) I e^cosx sinx dx

4) I x artctgx dx

5) I (1/x^2)*e^1/x dx

1) Divide the quadratic into the cubic and apply partial fraction decomposition to the remainder to get $x - 2 + \frac{5}{3(x+2)} - \frac{2}{3(x-1)}$ . Now integrate term-by-term.

2) Apply the identity $\sin A \sin B = \frac{1}{2} \left[ \cos ( A - B) - \cos (A + B) \right]$ . In your case A = 4x and B = 5x.

3) Make the substitution $u = \cos x$ .

4) Integration by parts.

5) Make the substitution $u = \frac{1}{x}$ .

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