The level of the numerator is much less than the level of the , as a result, the is factored, which ends up being the dice of an amount:

\$\$ x ^ 3 +3 x ^ 2 +3 x +1=( x +1) ^ 3\$\$

The essential is after that composed with the currently factored:

\$\$ intfrac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} dx= intfrac {3x +2} {(x +1) ^ 3} dx\$\$

Note that the contains a solitary straight variable that is elevated to the dice, whose kind is \$latex (px+ q) ^ m\$, with \$latex m= 3\$.

In such an instance, the disintegration right into basic portions of the integrand takes the list below kind:

\$\$ frac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} =frac {3x +2} {(x +1) ^ 3} =frac {} {x +1} +frac {B} {(x +1) ^ 2} +frac {C} {(x +1) ^ 3} \$\$

And currently we comply with a comparable procedure as in the previous instance, in order to locate the coefficients \$latex A\$, \$latex B\$ as well as \$latex C\$:

\$\$ frac {} {x +1} +frac {B} {(x +1) ^ 2} +frac {C} {(x +1) ^ 3} =frac {A( x +1) ^ 2+ B( x +1)+ C} {(x +1) ^ 3} \$\$

Then, the numerator is thoroughly created:

\$\$ A( x +1) ^ 2+ B( x +1)+ C =A( x ^ 2 +2 x +1)+ Bx+ B+C\$\$

\$\$ Ax ^ 2 +2 Ax+ A+B x+ B+C= Ax ^ 2+( 2A+ B) x+ B+C= 3x +2\$\$

By relating the coefficients of the equivalent powers on both sides of the equal rights, the complying with formulas are gotten:

\$ latex A= 0\$

\$ latex 2A+ B= 3Rightarrow B= 3\$

\$ latex B+C= 2Rightarrow C= -1\$

With these worths, the integrand is as adheres to:

\$\$ frac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} =frac {0} {x +1} +frac {3} {(x +1) ^ 2} -frac {1} {(x +1) ^ 3} \$\$

\$\$ frac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} =frac {3} {(x +1) ^ 2} -frac {1} {(x +1) ^ 3} \$\$

Now the initial essential is changed right into the amount of 2 quickly fixed integrals with an easy modification of variable:

\$\$ intfrac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} dx= 3intfrac {dx} {(x +1) ^ 2} -intfrac {dx} {(x +1) ^ 3} \$\$

The modification of variable is \$latex u =x +1\$, \$latex du= dx\$:

\$\$ intfrac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} dx= 3intfrac {du} {u ^ 2} -frac {du} {u ^ 3} \$\$

Applying the power regulation for assimilation, we acquire:

\$\$ intfrac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} dx= 3intfrac {du} {u ^ 2} -frac {du} {u ^ 3} \$\$

\$\$= 3int u ^ {-2} du-int u ^ {-3} du\$\$

\$\$= frac {3u ^ {-1}} {(-1)} -frac {u ^ {-2}} {(-2)} +C=- frac {3} {x +1} +frac {1} {2( x +1) ^ {2}} +C\$\$

Finally:

\$\$ intfrac {3x +2} {x ^ 3 +3 x ^ 2 +3 x +1} dx=- frac {3} {x +1} +frac {1} {2( x +1) ^ {2}} +C\$\$