Question

In the question given below find
$\int\limits_0^2 {{{\left( {{x^2} + 2x + 1} \right)}^{\dfrac{3}{2}}}\left( {x + 1} \right)dx = ?}$

Answer 1

The first bracket is the perfect square of a term .So substitute it by that term .Then using rules of indices we will cancel the powers. Then proceeding we will calculate the integrals.

Complete step-by-step answer:
Given that,
$\int\limits_0^2 {{{\left( {{x^2} + 2x + 1} \right)}^{\dfrac{3}{2}}}\left( {x + 1} \right)dx}$
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
So the first bracket is a perfect square of $x + 1$.
${\left( {x + 1} \right)^2} = {x^2} + 2x + 1$

$$\Rightarrow \int\limits_0^2 {{{\left( {{{\left( {x + 1} \right)}^2}} \right)}^{\dfrac{3}{2}}}\left( {x + 1} \right)dx} \\\ \Rightarrow \int\limits_0^2 {{{\left( {x + 1} \right)}^3}\left( {x + 1} \right)dx} \\\ \Rightarrow \int\limits_0^2 {{{\left( {x + 1} \right)}^{3 + 1}}dx} \\\ \Rightarrow \int\limits_0^2 {{{\left( {x + 1} \right)}^4}dx} \\\$$

Now finding the integrals using the formula,
$\int {{x^n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$

$$\Rightarrow \mathop {\left[ {\dfrac{{{{\left( {x + 1} \right)}^{4 + 1}}}}{{4 + 1}}} \right]}\nolimits_0^2 \\\ \Rightarrow \left[ {\dfrac{{{{\left( {x + 1} \right)}^5}}}{5}} \right]_0^2 \\\ \Rightarrow \dfrac{1}{5}\left[ {{{\left( {x + 1} \right)}^5}} \right]_0^2 \\\$$

Now apply the limits ,

$$\Rightarrow \dfrac{1}{5}\left[ {{{\left( {2 + 1} \right)}^5} - {{\left( {0 + 1} \right)}^5}} \right] \\\ \Rightarrow \dfrac{1}{5}\left[ {{3^5} - {1^5}} \right] \\\ \Rightarrow \dfrac{1}{5}\left[ {243 - 1} \right] \\\ \Rightarrow \dfrac{{242}}{5} \\\$$

$\int\limits_0^2 {{{\left( {{x^2} + 2x + 1} \right)}^{\dfrac{3}{2}}}\left( {x + 1} \right)dx} = \dfrac{{242}}{5}$

Additional information:
Here we have to perform integration but with limits to the function and have to find the value of the given function in between that limit.

Note: Don’t start directly with the integrals try to reduce it to the simplest form and then put up with the integrals. So that it reduces the time for calculations. Use the proper formulas for the integrals.