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Question: How do you use substitution to integrate \[\dfrac{x}{{\sqrt {x + 4} }}dx\] ?...

How do you use substitution to integrate xx+4dx\dfrac{x}{{\sqrt {x + 4} }}dx ?

Explanation

Solution

Hint : We need to evaluate ∫xx+4dx\int {\dfrac{x}{{\sqrt {x + 4} }}dx} . We know that the term inside the integral sign is called integrand. We simplify the integrand by making a substitution that is we put t=x+4t = x + 4 . After simplification we apply the integral. In the final we need the solution in terms of ‘x’ only. Here we have an indefinite integral.

Complete step by step solution:
Given,
∫xx+4dx\int {\dfrac{x}{{\sqrt {x + 4} }}dx}
to simplify the integrand let’s put t=x+4t = x + 4
Differentiating with respect to ‘x’ we have,
dt=dxdt = dx .
We have ‘x’ in the integrand t−4=xt - 4 = x .
Substituting these in the integral we have,
∫xx+4dx=∫(t−4)tdt\int {\dfrac{x}{{\sqrt {x + 4} }}dx} = \int {\dfrac{{(t - 4)}}{{\sqrt t }}dt}
=∫(t−4)tdt= \int {\dfrac{{(t - 4)}}{{\sqrt t }}dt}
=∫(t−4)(t)12dt= \int {\dfrac{{(t - 4)}}{{{{(t)}^{\dfrac{1}{2}}}}}dt} .
By using negative power rule of indices we have,
=∫(t−4)(t)−12dt= \int {(t - 4){{(t)}^{ - \dfrac{1}{2}}}dt}
=∫(t.t12−4t12)dt= \int {\left( {t.{t^{\dfrac{1}{2}}} - 4{t^{\dfrac{1}{2}}}} \right)dt}
=∫(t−12+1−4t−12)dt= \int {\left( {{t^{ - \dfrac{1}{2} + 1}} - 4{t^{ - \dfrac{1}{2}}}} \right)dt}
=∫(t12−4t−12)dt= \int {\left( {{t^{\dfrac{1}{2}}} - 4{t^{ - \dfrac{1}{2}}}} \right)dt}
We know that the differentiation of tn{t^n} with respect to ‘t’ is ∫tndt=tn+1n+1\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}}} ,
=(t12+1)(12+1)−4(t−12+1)(−12+1)+C= \dfrac{{\left( {{t^{\dfrac{1}{2} + 1}}} \right)}}{{\left( {\dfrac{1}{2} + 1} \right)}} - 4\dfrac{{\left( {{t^{ - \dfrac{1}{2} + 1}}} \right)}}{{\left( { - \dfrac{1}{2} + 1} \right)}} + C , where ‘C’ is the integration constant.
=(t32)(32)−4(t12)(12)+C= \dfrac{{\left( {{t^{\dfrac{3}{2}}}} \right)}}{{\left( {\dfrac{3}{2}} \right)}} - 4\dfrac{{\left( {{t^{\dfrac{1}{2}}}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} + C
=2(t32)3−4.(2)(t12)+C= \dfrac{{2\left( {{t^{\dfrac{3}{2}}}} \right)}}{3} - 4.(2)\left( {{t^{\dfrac{1}{2}}}} \right) + C
=2t323−8t12+C= \dfrac{{2{t^{\dfrac{3}{2}}}}}{3} - 8{t^{\dfrac{1}{2}}} + C
We need the answer in terms of x only, we have t=x+4t = x + 4 .
=2(x+4)323−8(x+4)12+C= \dfrac{{2{{\left( {x + 4} \right)}^{\dfrac{3}{2}}}}}{3} - 8{(x + 4)^{\dfrac{1}{2}}} + C
Thus we have, ∫xx+4dx=2(x+4)323−8(x+4)12+C\int {\dfrac{x}{{\sqrt {x + 4} }}dx} = \dfrac{{2{{\left( {x + 4} \right)}^{\dfrac{3}{2}}}}}{3} - 8{(x + 4)^{\dfrac{1}{2}}} + C , where ‘C’ is the integration constant.
So, the correct answer is “ ∫xx+4dx=2(x+4)323−8(x+4)12+C\int {\dfrac{x}{{\sqrt {x + 4} }}dx} = \dfrac{{2{{\left( {x + 4} \right)}^{\dfrac{3}{2}}}}}{3} - 8{(x + 4)^{\dfrac{1}{2}}} + C ”.

Note : Here we have an indefinite integral that is no upper limit and lower limit. Hence, in the case of indefinite integral we have integration constant. In definite integral we have lower limits and upper limits. Hence, in the case of definite integral we don’t have integration constant. As we can see in the above problem by using the substitute rule we can simplify the problem easily.