Answer
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Hint: For the given question we are given to solve the antiderivative of \[{{\left( 5x+1 \right)}^{2}}\]. For that let us expand the given equation by using basic algebraic formulas and then we have to integrate the resultant equation by using the basic integration formulas.
Complete step by step answer:
For the given problem we are given to find the antiderivative of the equation \[{{\left( 5x+1 \right)}^{2}}\].
Let us consider the given equation as equation (1) to get solved.
\[a={{\left( 5x+1 \right)}^{2}}...........\left( 1 \right)\]
For a polynomial of just degree 2, I would expand the equation and solve.
So, let us expand the equation (1) by using the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
Let us consider the given formula as formula (f1).
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}............\left( f1 \right)\]
By using the formula (f1) let us expand equation (1)
\[\Rightarrow a=25{{x}^{2}}+1+10x\]
Let us consider the above equation as equation (2).
\[\Rightarrow a=25{{x}^{2}}+1+10x................\left( 2 \right)\]
By doing antiderivative to the equation (2).
\[\Rightarrow a=\int{25{{x}^{2}}+1+10x\text{ dx}}\]
Let us consider the given equation as equation (2).
\[\Rightarrow a=\int{25{{x}^{2}}+1+10x\text{ dx}}...............\left( 2 \right)\]
As we know the formula \[\int{{{a}^{x}}dx=\dfrac{{{a}^{x+1}}}{x+1}+c}\].
Let us consider the above formula as (f2).
\[\int{{{a}^{x}}dx=\dfrac{{{a}^{x+1}}}{x+1}+c}................\left( 2 \right)\]
By applying formula (f2) to equation (2), we get
\[\Rightarrow a=25\dfrac{{{x}^{3}}}{3}+x+10\dfrac{{{x}^{2}}}{2}\]
By simplifying a bit, we get
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+x+5{{x}^{2}}+c\]
By arranging terms in power wise, we get
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+5{{x}^{2}}+x+c\]
Therefore, let us consider the equation as equation (3).
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+x+5{{x}^{2}}+c...............\left( 3 \right)\]
Therefore, antiderivative of \[{{\left( 5x+1 \right)}^{2}}\] is \[a=\dfrac{25}{3}{{x}^{3}}+5{{x}^{2}}+x+c\].
Note:
We should know that antiderivative means summation i.e. integration. We can do this problem by another way i.e. we can do direct integration to the given equation without expanding. But I think the way I solved this problem is easier than any other method.
Complete step by step answer:
For the given problem we are given to find the antiderivative of the equation \[{{\left( 5x+1 \right)}^{2}}\].
Let us consider the given equation as equation (1) to get solved.
\[a={{\left( 5x+1 \right)}^{2}}...........\left( 1 \right)\]
For a polynomial of just degree 2, I would expand the equation and solve.
So, let us expand the equation (1) by using the formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\].
Let us consider the given formula as formula (f1).
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}............\left( f1 \right)\]
By using the formula (f1) let us expand equation (1)
\[\Rightarrow a=25{{x}^{2}}+1+10x\]
Let us consider the above equation as equation (2).
\[\Rightarrow a=25{{x}^{2}}+1+10x................\left( 2 \right)\]
By doing antiderivative to the equation (2).
\[\Rightarrow a=\int{25{{x}^{2}}+1+10x\text{ dx}}\]
Let us consider the given equation as equation (2).
\[\Rightarrow a=\int{25{{x}^{2}}+1+10x\text{ dx}}...............\left( 2 \right)\]
As we know the formula \[\int{{{a}^{x}}dx=\dfrac{{{a}^{x+1}}}{x+1}+c}\].
Let us consider the above formula as (f2).
\[\int{{{a}^{x}}dx=\dfrac{{{a}^{x+1}}}{x+1}+c}................\left( 2 \right)\]
By applying formula (f2) to equation (2), we get
\[\Rightarrow a=25\dfrac{{{x}^{3}}}{3}+x+10\dfrac{{{x}^{2}}}{2}\]
By simplifying a bit, we get
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+x+5{{x}^{2}}+c\]
By arranging terms in power wise, we get
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+5{{x}^{2}}+x+c\]
Therefore, let us consider the equation as equation (3).
\[\Rightarrow a=\dfrac{25}{3}{{x}^{3}}+x+5{{x}^{2}}+c...............\left( 3 \right)\]
Therefore, antiderivative of \[{{\left( 5x+1 \right)}^{2}}\] is \[a=\dfrac{25}{3}{{x}^{3}}+5{{x}^{2}}+x+c\].
Note:
We should know that antiderivative means summation i.e. integration. We can do this problem by another way i.e. we can do direct integration to the given equation without expanding. But I think the way I solved this problem is easier than any other method.
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