Question

How do you find the integral of$\left( {\left( x \right)\sqrt {x - 1} } \right)dx$ ?

Answer 1

Indefinite integral simply represents the area under a given curve without any boundary conditions. So here by using this basic definition we can integrate $\left( {\left( x \right)\sqrt {x - 1} } \right)dx$. Also we know one of the basic identity:$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$. The above expression and equation can be used to integrate $\left( {\left( x \right)\sqrt {x - 1} } \right)dx$.

Complete step by step answer:
Given, $\left( {\left( x \right)\sqrt {x - 1} } \right)dx.............................................\left( i \right)$
Also by the basic definition of indefinite integral we can write that:
Indefinite integral is given by: $\int {f\left( x \right)dx}$
Such to integrate $\left( {\left( x \right)\sqrt {x - 1} } \right)dx$ we can write
$\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} ..........................\left( {ii} \right)$
Now on observing (ii) we can say that the term $\left( {\left( x \right)\sqrt {x - 1} } \right)dx$ cannot be integrated directly such that let’s assume:
$x - 1 = t................................\left( {iii} \right) \\\ \Rightarrow x = t + 1 \\\$
Now let’s differentiate equation (ii) and find the value of $dx$ such that we can substitute it in (i):
So we get:

\Rightarrow dx = dt......................\left( {iv} \right) \\\ $$ Now let’s substitute all the above values in (i): So we can write: $$\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {t + 1} \right)\sqrt t dt} ..........................\left( v \right)$$ We know one of the exponential property that: ${x^m} \times {x^n} = {x^{m + n}}$ So we can apply it to (v) and we can write: $$\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {t + 1} \right)\sqrt t dt} \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {t\sqrt t + \sqrt t } \right)dt} \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {{t^{1 + \dfrac{1}{2}}} + {t^{\dfrac{1}{2}}}} \right)dt} \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {{t^{\dfrac{3}{2}}} + {t^{\dfrac{1}{2}}}} \right)dt} ...............\left( {vi} \right) \\\ $$ Also then to integrate (vi) we can use one of the basic identity which is: $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $ Here instead of $x$ we have $t$. So we can write: $$\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \int {\left( {{t^{\dfrac{3}{2}}} + {t^{\dfrac{1}{2}}}} \right)dt} \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \dfrac{{{t^{\dfrac{3}{2} + 1}}}}{{\left( {\dfrac{3}{2} + 1} \right)}} + \dfrac{{{t^{\dfrac{1}{2} + 1}}}}{{\left( {\dfrac{1}{2} + 1} \right)}} + C \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \dfrac{{{t^{\dfrac{5}{2}}}}}{{\left( {\dfrac{5}{2}} \right)}} + \dfrac{{{t^{\dfrac{3}{2}}}}}{{\left( {\dfrac{3}{2}} \right)}} + C \\\ \Rightarrow\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \dfrac{2}{5}{t^{\dfrac{5}{2}}} + \dfrac{2}{3}{t^{\dfrac{3}{2}}} + C........................\left( {vii} \right) \\\ $$ Now let’s substitute the value of $t$ back in the equation (vii) such that we can write: $$\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx} = \dfrac{2}{5}{t^{\dfrac{5}{2}}} + \dfrac{2}{3}{t^{\dfrac{3}{2}}} + C \\\ \therefore\int {\left( {\left( x \right)\sqrt {x - 1} } \right)dx}= \dfrac{2}{5}{\left( {x - 1} \right)^{\dfrac{5}{2}}} + \dfrac{2}{3}{\left( {x - 1} \right)^{\dfrac{3}{2}}} + C.......................\left( {viii} \right) \\\ $$ **Therefore from (viii) we can write that on integrating $\left( {\left( x \right)\sqrt {x - 1} } \right)dx$ we get $$\dfrac{2}{5}{\left( {x - 1} \right)^{\dfrac{5}{2}}} + \dfrac{2}{3}{\left( {x - 1} \right)^{\dfrac{3}{2}}} + C$$.** **Note:** Since the basic definition indefinite integral simply implies the area under a curve such that the value of an integral must be finite or else the integral doesn’t exist. Also if we cannot integrate an expression directly then we have to use the formula for integration by parts which is given by: $\int {udv = uv - \int {vdu} } $