How do you integrate $\int \frac{x^{3} e x^{2}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3ex^2)/(x^2 +1)^2dx$ ?

Question

How do you integrate $\int \frac{x^{3} e x^{2}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3ex^2)/(x^2 +1)^2dx$ ?

$\int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3e^(x^2))/(x^2 +1)^2dx$ ?

2 Answers

Impact of this question

17125 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-12T19:39:33

$\int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x = \frac{e^{x^{2}}}{2 (x^{2} + 1)} + c$ $int(x^3e^(x^2))/(x^2+1)^2dx=e^(x^2)/(2(x^2+1))+c$

Explanation:

Here,

$I = \int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x$ $I=int(x^3e^(x^2))/(x^2+1)^2dx$

$= \int \frac{x^{2} e^{x^{2}}}{{(x^{2} + 1)}^{2}} \cdot x d x$ $=int(x^2e^(x^2))/(x^2+1)^2*xdx$

Subst . $x^{2} = u \Rightarrow 2 x d x = d u \Rightarrow x d x = \frac{1}{2} d u$ $color(violet)(x^2=u=>2xdx=du=>xdx=1/2du$

So,

$I = \int \frac{u e^{u}}{{(u + 1)}^{2}} \cdot \frac{1}{2} d u$ $I=int(ue^u)/(u+1)^2*1/2du$

$= \frac{1}{2} \int \frac{(u + 1 - 1) e^{u}}{{(u + 1)}^{2}} d u$ $=1/2int((u+1-1)e^u)/(u+1)^2du$

$= \frac{1}{2} \int {\frac{(u + 1)}{{(u + 1)}^{2}} - \frac{1}{{(u + 1)}^{2}}} e^{u} d u$ $=1/2int{((u+1))/(u+1)^2-1/(u+1)^2}e^udu$

$= \frac{1}{2} {\int \frac{1}{u + 1} e^{u} d u - \int \frac{1}{{(u + 1)}^{2}} e^{u} d u}$ $=1/2{color(blue)(int1/(u+1)e^udu)-color(red) (int1/(u+1)^2e^udu)}$

Using Integration by parts: in the first integral

$I$ $I$ = $\frac{1}{2} {[\frac{1}{u + 1} \int e^{u} - \int \frac{- 1}{{(u + 1)}^{2}} e^{u} d u] - \int \frac{1}{{(u + 1)}^{2}} e^{u} d u}$ $1/2{color(blue)([1/(u+1)inte^u-int(-1)/(u+1)^2e^udu])- color(red)(int1/(u+1)^2e^udu)}$

$I$ $I$ = $\frac{1}{2} {\frac{1}{u + 1} e^{u} + \int \frac{1}{{(u + 1)}^{2}} e^{u} d u - \int \frac{1}{{(u + 1)}^{2}} e^{u} d u$ $1/2{color(brown)(1/(u+1)e^u+color(blue) (int(1)/(u+1)^2e^udu)-color(red)(int1/(u+1)^2e^udu)}$

$I = \frac{1}{2} \cdot \frac{1}{u + 1} e^{u} + c$ $I=1/2 *color(brown)(1/(u+1)e^u)+c$

Subst. back $u = x^{2}$ $color(violet)( u=x^2$ we get

$I = \frac{1}{2} (\frac{1}{x^{2} + 1}) e^{x^{2}} + c$ $I=1/2(1/(x^2+1))e^(x^2)+c$

Hence ,

$I = \frac{e^{x^{2}}}{2 (x^{2} + 1)} + c$ $I=e^(x^2)/(2(x^2+1))+c$
.....................................................................................

Note: Change in the question is as below.

$x^{3} e x^{2} \to x^{3} e^{x^{2}}$ $x^3ex^2tox^3e^(x^2)$

Answer 2 · 2018-07-12T19:48:19

The answer is $= \frac{e^{x^{2}}}{2 (x^{2} + 1)} + C$ $=(e^(x^2))/(2(x^2+1))+C$

Explanation:

Perform the substitution

$u = x^{2}$ $u=x^2$ , $\Rightarrow$ $=>$ , $d u = 2 x d x$ $du=2xdx$

The integral is

$I = \int \frac{x^{3} e^{x^{2}} d x}{{(x^{2} + 1)}^{2}}$ $I=int(x^3e^(x^2)dx)/(x^2+1)^2$

$= \frac{1}{2} \int \frac{u e^{u} d u}{{(u + 1)}^{2}}$ $=1/2int(ue^udu)/(u+1)^2$

Perform an integration by parts

$intwv'dx=wv-intw'vdx$

$w=ue^u$ , $=>$ , $w'=ue^u +e^u=e^u(u+1)$

$v'=1/(u+1)^2$ , $=>$ , $v=-1/(u+1)$

Therefore,

$I=-1/2(ue^u)/(u+1)+1/2inte^udu$

$=e^u/2-(ue^u)/(2(u+1))$

$=e^(x^2)/2-(x^2e^(x^2))/(2(x^2+1))+C$

$=1/2e^(x^2)((x^2+1-x^2))/(x^2+1)+C$

$=(e^(x^2))/(2(x^2+1))+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{x^{3} e x^{2}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3ex^2)/(x^2 +1)^2dx$ ?

$\int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3e^(x^2))/(x^2 +1)^2dx$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫(x^3ex^2)/(x^2 +1)^2dx∫x3ex2(x2+1)2dx∫(x^3ex^2)/(x^2 +1)^2dx?

∫(x^3e^(x^2))/(x^2 +1)^2dx ∫x3ex2(x2+1)2dx∫(x^3e^(x^2))/(x^2 +1)^2dx ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{x^{3} e x^{2}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3ex^2)/(x^2 +1)^2dx$ ?

$\int \frac{x^{3} e^{x^{2}}}{{(x^{2} + 1)}^{2}} d x$ $∫(x^3e^(x^2))/(x^2 +1)^2dx$ ?