Question

Evaluate the integration \[\int {\dfrac{{{x^3} + 3{x^2} + 3x + 1}}{{{{\left( {x + 1} \right)}^5}}}} dx\].
A. \[ - \dfrac{1}{{x + 1}} + c\]
B. \[\dfrac{1}{5}\log \left( {x + 1} \right) + c\]
C. \[\log \left( {x + 1} \right) + c\]
D. \[{\tan ^{ - 1}}x + c\]

Answer 1

Hint: First we apply the formula \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\] in the numerator of the expression \[\dfrac{{{x^3} + 3{x^2} + 3x + 1}}{{{{\left( {x + 1} \right)}^5}}}\]. Then cancel out common terms and apply the integration formula \[\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c\].

Formula Used:
\[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\]
\[\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c\]

Complete step by step solution:
Given integration is \[\int {\dfrac{{{x^3} + 3{x^2} + 3x + 1}}{{{{\left( {x + 1} \right)}^5}}}} dx\].
Apply the formula \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\] in the numerator
\[ = \int {\dfrac{{{{\left( {x + 1} \right)}^3}}}{{{{\left( {x + 1} \right)}^5}}}} dx\]
Cancel out \[{\left( {x + 1} \right)^3}\]
\[ = \int {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} dx\]
Now applying the formula \[\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c\], since the coefficient of \[x\] is 1.
\[ = - \dfrac{1}{{\left( {x + 1} \right)}} + c\] where \[c\] is an integrating factor

Therefore option A is the correct option

Note:Sometimes students use substitute method to solve the integration \[\int {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} dx\]. If the coefficient of \[x\] is 1 then we can apply the formula \[\int {\dfrac{1}{{{x^2}}}} dx = - \dfrac{1}{x} + c\] directly. If the coefficient of \[x\] is not 1, then we have to apply the substitute method.