(x²+1 ex dx =
e
(x + 1)²
Answers
Answer:
I &= \int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx
To Find:
We have to integrate the given function.
Solution:
Let I &= \int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx
Adding and subtracting 2x2x
I&= \int \frac{e^x \left(\left( x + 1 \right)^2 - 2x\right)}{\left( x + 1 \right)^2} \,\mathrm dx
\begin{gathered}&= \int \frac{e^x \left( x + 1 \right)^2}{\left( x + 1 \right)^2} \,\mathrm dx - 2 \int \frac{e^x\left( x + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx + 2 \int \frac{e^x}{\left( x + 1 \right)^2} \,\mathrm dx \\ \\&= \int e^x \,\mathrm dx - 2 \int \frac{e^x}{x + 1} \,\mathrm dx + 2 \int \frac{e^x}{\left( x + 1 \right)^2} \,\mathrm dx \\\end{gathered}
=∫
(x+1)
2
e
x
(x+1)
2
dx−2∫
(x+1)
2
e
x
(x+1)
dx+2∫
(x+1)
2
e
x
dx
=∫e
x
dx−2∫
x+1
e
x
dx+2∫
(x+1)
2
e
x
dx
I &= e^x - 2 \left( \int \frac{e^x}{x + 1} \,\mathrm dx - \int \frac{e^x}{\left( x + 1 \right)^2} \,\mathrm dx \right) \end{aligned}
We use integration by parts on the first of these integrals, which uses the following general rule:
\begin{gathered}[\int u \,\mathrm dv = uv - \int v du\\u = \frac{1}{x+1} \\v =e^x\end{gathered}
[∫udv=uv−∫vdu
u=
x+1
1
v=e
x
\therefore \displaystyle \int \frac{e^x}{x+1} \,\mathrm dx = \frac{e^x}{x+1} + \int \frac{e^x}{\left(x+1\right)^2} \,\mathrm dx∴∫
x+1
e
x
dx=
x+1
e
x
+∫
(x+1)
2
e
x
dx
\begin{gathered}I&= e^x - 2 \left( \frac{e^x}{x+1} + \int \frac{e^x}{\left(x+1\right)^2} \,\mathrm dx - \int \frac{e^x}{\left( x + 1 \right)^2} \,\mathrm dx \right) \\\end{gathered}
I
=e
x
−2(
x+1
e
x
+∫
(x+1)
2
e
x
dx−∫
(x+1)
2
e
x
dx)
\begin{gathered}&= e^x - \frac{2e^x}{x+1} \\\\ &= \frac{e^x\left(x+1\right) - 2e^x}{x+1} \\\end{gathered}
=e
x
−
x+1
2e
x
=
x+1
e
x
(x+1)−2e
x
I&= \boxed{ \frac{e^x\left(x-1\right)}{x+1} + C } \end{aligned}
Thus,
\int \frac{e^x \left( x^2 + 1 \right)}{\left( x + 1 \right)^2} \,\mathrm dx &= \boxed{ \frac{e^x\left(x-1\right)}{x+1} + C } \end{aligned}