Question

Solve $\int {\dfrac{{{x^3}}}{{{{({x^2} + 1)}^3}}}} dx$

Answer 1

Hint: We can solve this integration by simply assuming $({x^2} + 1)$as $t$and then converting the whole integration in terms of $t$. Also, ${x^2} + 1 = t \Rightarrow 2xdx = dt$. The numerator can be written as ${x^2}.xdx$. Replace ${x^2}$ by ($t - 1$) and $xdx$by $\dfrac{{dt}}{2}$. In this way, the numerator and denominator will be reduced in terms of $t$which we can solve easily.

Complete Step by Step Solution:
Suppose $I = \int {\dfrac{{{x^3}}}{{{{({x^2} + 1)}^3}}}} dx................$(equation $1$)
Let ${x^2} + 1 = t........$(equation $2$)
On differentiating equation$2$,
$
   \Rightarrow 2xdx = dt \\
   \Rightarrow dx = \dfrac{{dt}}{2} \\
 $
We can rewrite equation $1$as $I = \int {\dfrac{{{x^2}.x}}{{{{({x^2} + 1)}^3}}}} dx$
Replacing ${x^2}$ by ($t - 1$) and $xdx$by $\dfrac{{dt}}{2}$,
$ \Rightarrow I = \int {\dfrac{{(t - 1)}}{{{{(t)}^3}}}} \dfrac{{dt}}{2}$
Taking $\dfrac{1}{2}$outside the integration, as it is a constant;
$
   \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{(t - 1)}}{{{{(t)}^3}}}} dt \\
   \Rightarrow I = \dfrac{1}{2}\int {(\dfrac{1}{{{t^2}}}} - \dfrac{1}{{{t^3}}})dt \\
 $
Integration of ${t^n}$is $\dfrac{{{t^{n + 1}}}}{{n + 1}}$.
$
    \\
   \Rightarrow I = \dfrac{1}{2}\int {({t^{ - 2}} + {t^{ - 3}})} dt \\
   \Rightarrow I = \dfrac{1}{2}(\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + \dfrac{{{t^{ - 3 + 1}}}}{{ - 3 + 1}} + C) \\
   \Rightarrow I = - \dfrac{1}{{2t}} - \dfrac{1}{{4{t^2}}} + {C_1} \\
    \\
 $
Substituting the value of $t$ from equation $2$,
$I = - \dfrac{1}{{2({x^2} + 1)}} - \dfrac{1}{{4{{({x^2} + 1)}^2}}} + {C_1}$

Note:
We can also solve this question like this;
Let $x = \tan \theta $
$ \Rightarrow dx = {\sec ^2}\theta d\theta $
$ \Rightarrow I = \int {\dfrac{{{{\tan }^3}\theta }}{{{{(1 + {{\tan }^2}\theta )}^3}}}} {\sec ^2}\theta d\theta $
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
$ \Rightarrow I = \int {\dfrac{{{{\tan }^3}\theta }}{{{{\sec }^6}\theta }}} {\sec ^2}\theta d\theta $
$ \Rightarrow I = \int {\dfrac{{{{\tan }^3}\theta }}{{{{\sec }^4}\theta }}} d\theta $
$ \Rightarrow I = \int {\dfrac{{{{\sin }^3}\theta }}{{{{\cos }^3}\theta }}} \times {\cos ^4}\theta d\theta $
$ \Rightarrow I = \int {{{\sin }^3}} \theta \times \cos \theta d\theta $
Let $\sin \theta = t$$ \Rightarrow \cos \theta d\theta = dt$
Then $I = \int {{t^3}} dt$
$ \Rightarrow I = \dfrac{{{t^4}}}{4} + C$ As, $t = \sin \theta $$ \Rightarrow I = \dfrac{{{{\sin }^4}\theta }}{4} + C$
Now, we have $x = \tan \theta $$ \Rightarrow \sin \theta = \dfrac{x}{{\sqrt {{{(1 + x)}^2}} }}$
Putting this in place of $\sin \theta $,we will get the value of $I$.
It is possible that we get different answers by different methods but if you will solve properly, then on differentiating the result, we will end up getting the same value. Remember important formulas of integration.