Find integral of $\frac{((x^{2}) + 1) . e^{x}}{{(x + 1)}^{2}}$ $(((x^2)+1).e^x)/(x+1)^2$ With respect to $x$ $x$ ?

Question

4022 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T07:49:11

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

$\Rightarrow \int e^{x} [\frac{((x^{2} - 1) + 2)}{{(x + 1)}^{2}}] d x$ $=>inte^x[(((x^2-1)+2))/(x+1)^2] dx$

$\Rightarrow \int e^{x} [\frac{(x - 1) (x + 1) + 2}{{(x + 1)}^{2}}] d x$ $=>inte^x[((x-1)(x+1)+2)/(x+1)^2] dx$

$\Rightarrow \int e^{x} [\frac{x - 1}{x + 1}] d x + 2 \int e^{x} [\frac{1}{{(x + 1)}^{2}}] d x$ $=>inte^x[(x-1)/(x+1)]dx+2inte^x[1/(x+1)^2]dx$

Notice,
Considering $f (x) = [\frac{x - 1}{x + 1}]$ $f(x) = [(x-1)/(x+1)]$

$f'(x) = ((x+1)(x-1)'-(x-1)(x+1)')/(x+1)^2$

$=> ((x+1)-(x-1))/(x+1)^2 => 2/(x+1)^2$

So the integral is now of the form,

$=>inte^x[f(x)]dx+inte^x[f'(x)]dx$

$=> f(x) + c$

$=> e^x[(x-1)/(x+1)] + c$

Find integral of (((x^2)+1).e^x)/(x+1)^2 ((x2)+1).ex(x+1)2(((x^2)+1).e^x)/(x+1)^2 With respect to xxx?