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I'm trying to solve this indefinite integral by means of partial fraction decomposition:

$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.

The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the integrand function in this form:

$\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{Ax+B}{\left(x^2+4x+5\right)}+\dfrac{Cx+D}{\left(x^2+4x+5\right)^2}$.

I get:

$\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{\left(Ax+B\right)\left(x^2+4x+5\right)+Cx+D}{\left(x^2+4x+5\right)^2}$.

Multiplying the right term:

$\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{Ax^3+4Ax^2+5Ax+Bx^2+4Bx+5B+Cx+D}{\left(x^2+4x+5\right)^2}$.

Now i collect the terms with the same pwer of $x$:

$\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D+ 5B}{\left(x^2+4x+5\right)^2}$.

Now i equate the two numerators:

$x+1=Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D$

and equate term by term: i get:

$A=0$, $B=0$, $C=1$, $D=1$.

With these values i get a correct identity:

$\dfrac{x+1}{\left(x^2+4x+5\right)^2}= \dfrac{x+1}{\left(x^2+4x+5\right)^2}$

but this is unuseful in order to solve the integral.

Where is my mistake ?

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  • $\begingroup$ If partial fraction decomposition is not mandatory, choose $x+2=\tan y$ $\endgroup$ – lab bhattacharjee Jul 29 '16 at 16:12
  • $\begingroup$ You are missing a $5B$ in the numerator. I inserted three stars $***$ for you. That is where the $5B$ has gone missing $\endgroup$ – imranfat Jul 29 '16 at 16:23
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Here is an easy way $$\int { \frac { x+1 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\frac { 1 }{ 2 } \int { \frac { 2x+4-2 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\\ =\frac { 1 }{ 2 } \int { \frac { 2x+4 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } dx-\int { \frac { dx }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } } =\\ =\frac { 1 }{ 2 } \underset { { I }_{ 1 } }{ \underbrace { \int { \frac { d\left( x^{ 2 }+4x+5 \right) }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } } } -\underset { { I }_{ 2 } }{ \underbrace { \int { \frac { d\left( x+2 \right) }{ { \left( { \left( x+2 \right) }^{ 2 }+1 \right) }^{ 2 } } } } } =$$ Obviously,$$\\ { I }_{ 1 }=-\frac { 1 }{ 2\left( x^{ 2 }+4x+5 \right) } ,$$ now,let calculate ${ I }_{ 2 }$,substitute here $x+2=\tan { z } $,so that $${ I }_{ 2 }=\int { \frac { d\tan { z } }{ { \left( { \tan ^{ 2 }{ z } }+1 \right) }^{ 2 } } } =\int { \frac { \cos ^{ 4 }{ z } }{ \cos ^{ 2 }{ z } } } dz=\int { \cos ^{ 2 }{ z } =\frac { 1 }{ 2 } \int { \left( 1+\cos { 2z } \right ) dz } } =\frac { 1 }{ 2 } \left( z+\frac { \sin { 2z } }{ 2 } \right) $$ finally,

$$\int { \frac { x+1 }{ \left( x^{ 2 }+4x+5 \right) ^{ 2 } } } dx=\frac { 1 }{ 2 } \left( -\frac { 1 }{ \left( x^{ 2 }+4x+5 \right) } -\arctan { \left( x+2 \right) -\frac { \sin { 2\left( \arctan { \left( x+2 \right) } \right) } }{ 2 } } \right) +C$$

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    $\begingroup$ much better then partial fractions $+1$ $\endgroup$ – tired Jul 29 '16 at 16:51
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As suggested in the comment directly use Trigonometric substitution

As $x^2+4x+5=(x+2)^2+1^2,$ let $\arctan(x+2)=y\implies x+2=\tan y$

$$2I\int\dfrac{x+1}{(x^2+4x+5)^2}dx=\int(\tan y-1)\cos^2y\ dy$$

$$4I=2\int(\sin2y-1-\cos2y)dy=-\sin2y-2y-\cos2y+K$$

Now $\sin2y=\dfrac{2\tan y}{1+\tan^2y}=\cdots$

and $\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac2{1+\tan^2y}-1=\cdots$

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  • $\begingroup$ I liked Battani's approach because I thought that he would avoid conventional wisdom, which is a trig sub. But then he used a trig sub nonetheless. Might as well go by your sledgehammer trig sub, which is probably what the majority of my students would do...+1 $\endgroup$ – imranfat Jul 29 '16 at 20:44
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Note that we can write the partial fraction expansion of $\frac{1}{(x+4x+5)^2}$ as

$$\frac{1}{(x+4x+5)^2}=\frac{1}{(x-r)^2(x-r^*)^2}=\frac{A}{x-r}+\frac{B}{(x-r)^2}+\frac{C}{x-r^*}+\frac{D}{(x-r^*)^2}$$

where $r=-2+i$ and $r^*=-2- i$.

Multiplying by $(x-r)^2$ and letting $x\to r$ reveals that $B=\frac{1}{(r-r^*)^2}=-\frac14$.

Multiplying by $(x-r)^2$, taking a derivative with respect to $x$, and letting $x\to r$ reveals that $A=-\frac{2}{(r-r^*)^3}=-\frac{i}{4}$.

Multiplying by $(x-r^*)^2$ and letting $x\to r^*$ reveals that $D=\frac{1}{(r^*-r)^2}=-\frac14$.

Multiplying by $(x-r^*)^2$, taking a derivative with respect to $x$, and letting $x\to r^*$ reveals that $C=\frac{2}{(r-r^*)^3}=\frac{i}{4}$.


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  • $\begingroup$ $C=\frac{i}{4}$ Heelp:) $\endgroup$ – imranfat Jul 29 '16 at 16:36
  • $\begingroup$ @imranfat I don't understand your comment. $\endgroup$ – Mark Viola Jul 29 '16 at 17:06
  • $\begingroup$ MV Is the $C$ value really supposed to be complex? $\endgroup$ – imranfat Jul 29 '16 at 17:46
  • $\begingroup$ @imranfat Yes, $A$ and $C$ are purely imaginary. $\endgroup$ – Mark Viola Jul 29 '16 at 17:53
  • $\begingroup$ That means, that if we only consider real numbers, PFD cannot be performed? At least not according to quickmath.com/webMathematica3/quickmath/algebra/… $\endgroup$ – imranfat Jul 29 '16 at 18:00

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