How do you evaluate the definite integral $\int \frac{2}{{(x + 1)}^{2}}$ $int 2/(x+1)^2$ from $[0, 3]$ $[0,3]$ ?

Question

1574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-30T17:26:00

$\int_{0}^{3} \frac{2}{{(x + 1)}^{2}} d x = \frac{3}{2}$ $int_0^3 2/(x+1)^2dx=3/2$

Use the substitution method as follows :

Let $u = x + 1$ $u=x+1$

$therefore du=dx$

The limits of integration will then change as follows :

$x=0 => u=1 and x=3 => u = 4$

Hence the original integral may be written as

$int_0^3 2/(x+1)^2dx=2 int_1^4 1/u^2 du$

$=-2[1/u]_1^4$

$=-2(1/4-1)$

$=3/2$

How do you evaluate the definite integral int 2/(x+1)^2∫2(x+1)2int 2/(x+1)^2 from [0,3][0,3][0,3]?