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Math Help - Help on indefinite integration needed!

  1. #1
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    Exclamation Help on indefinite integration needed!

    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



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  2. #2
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    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

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  3. #3
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    Quote Originally Posted by lightbird View Post
    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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