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Question: -What is the integral of \[\int {\dfrac{x}{{{{\left( {x + 1} \right)}^3}}}dx} \]?...

-What is the integral of ∫x(x+1)3dx\int {\dfrac{x}{{{{\left( {x + 1} \right)}^3}}}dx} ?

Explanation

Solution

Hint : To solve this type of question we use concepts of integration without limits .Let us substitute in the given integral to find the value of the given integral easily. So, let us assume the denominator as t and then substitute the numerator in terms of dt.

Complete step-by-step answer :
Now, we will use the substitution technique. We will let the denominator term as t and then differentiate the denominator term because the given integral is in proper form i.e. the degree of numerator is less than degree of denominator. So,
Let x+1=tx + 1 = t
Differentiating both sides with respect to x, we get
d(x)dx+d(1)dx=dtdx\dfrac{{d(x)}}{{dx}} + \dfrac{{d(1)}}{{dx}} = \dfrac{{dt}}{{dx}}
1=dtdx⇒dt=dx1 = \dfrac{{dt}}{{dx}} \Rightarrow dt = dx
(As we know d(xn)dx=nxn−1\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}} n=1n = 1 for d(x)dx=x1−1=x0=1\dfrac{{d(x)}}{{dx}} = {x^{1 - 1}} = {x^0} = 1 and for other term on L.H.S \dfrac{{d(1)}}{{dx}} \Rightarrow n = 0$$$$ \Rightarrow \dfrac{{d(1)}}{{dx}} = 0)
Substituting the value of t and dx in the given integral and t=x+1⇒x=t−1t = x + 1 \Rightarrow x = t - 1 we get,
∫x(x+1)3dx=∫t−1(t)3dt\int {\dfrac{x}{{{{\left( {x + 1} \right)}^3}}}dx} = \int {\dfrac{{t - 1}}{{{{\left( t \right)}^3}}}dt}
Now splitting this we can write as follows
∫(tt3−1t3)dt=∫(1t2−1t3)dt=∫(t−2−t−3)dt=∫t−2dt−∫t−3dt\int {(\dfrac{t}{{{t^3}}} - \dfrac{1}{{{t^3}}})dt = } \int {(\dfrac{1}{{{t^2}}} - \dfrac{1}{{{t^3}}})dt = } \int {({t^{ - 2}} - {t^{ - 3}})dt = } \int {{t^{ - 2}}dt - \int {{t^{ - 3}}dt} }
Now as we use the formula
∫xndx=xnn+1+c\int {{x^n}dx} = \dfrac{{{x^n}}}{{n + 1}} + c where cc is the integration constant
\Rightarrow $$$$\int {{t^{ - 2}}dt - \int {{t^{ - 3}}dt} } =t−2−2+1+c1+t−3−3+1+c2 = \dfrac{{{t^{ - 2}}}}{{ - 2 + 1}} + {c_1} + \dfrac{{{t^{ - 3}}}}{{ - 3 + 1}} + {c_2}
=−t−21−t−32+c2+c1= - \dfrac{{{t^{ - 2}}}}{1} - \dfrac{{{t^{ - 3}}}}{2} + {c_2} + {c_1}
=−(2t+12t3)+c2+c1= - \left( {\dfrac{{2t + 1}}{{2{t^3}}}} \right) + {c_2} + {c_1}
Let us take c2+c1=c{c_2} + {c_1} = c
We get ∫t−1(t)3dt=−(2t+12t3)+c\int {\dfrac{{t - 1}}{{{{\left( t \right)}^3}}}dt} = - \left( {\dfrac{{2t + 1}}{{2{t^3}}}} \right) + c
Now putting the value of t in the above equation, we get
−(2t+12t3)+c=−(2(x+1)+12(x+1)3)+c=−12(2x+3(x+1)3)+c- \left( {\dfrac{{2t + 1}}{{2{t^3}}}} \right) + c = - \left( {\dfrac{{2\left( {x + 1} \right) + 1}}{{2{{\left( {x + 1} \right)}^3}}}} \right) + c = - \dfrac{1}{2}\left( {\dfrac{{2x + 3}}{{{{\left( {x + 1} \right)}^3}}}} \right) + c
So, the correct answer is “−12(2x+3(x+1)3)+c - \dfrac{1}{2}\left( {\dfrac{{2x + 3}}{{{{\left( {x + 1} \right)}^3}}}} \right) + c”.

Note : While solving questions which include integration of given terms, we have to check whether the given integral is proper or improper. In a proper integral, the degree of the numerator is less than that of the denominator and vice – versa in the improper integral. Also, we have to write the integration constant c when we are dealing with indefinite integrals i.e. integrals with no limit.