Chapter 25, Problem 1RE

To determine

Whether the equation “${\displaystyle \int {\left(3x^2+1\right)}^{5}dx=\frac{1}{6}{\left(3x^2+1\right)}^6}+C$” is true or false.

Answer to Problem 1RE

The equation “${\displaystyle \int {\left(3x^2+1\right)}^{5}dx=\frac{1}{6}{\left(3x^2+1\right)}^6}+C$” is $\underset{\_}{\text{false}}$.

Explanation of Solution

Given that the equation is ${\displaystyle \int {\left(3x^2+1\right)}^{5}dx=\frac{1}{6}{\left(3x^2+1\right)}^6}+C$.

Evaluate the left hand side of the given equation as,

$\begin{array}{c}{\displaystyle \int {\left(3x^2+1\right)}^{5}dx}={\displaystyle \int \left(243x^{10}+405x^8+270x^6+90x^4+15x^2+1\right)dx}\\ =\frac{243x^{11}}{11}+\frac{405x^9}{9}+\frac{270x^7}{7}+\frac{90x^5}{5}+\frac{15x^3}{3}+x+C\mathrm{...}\left(1\right)\end{array}$

Expand the expression for right hand side using the formula for binomial expansion ${\left(a+b\right)}^n={\displaystyle \sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)}{a}^{n-k}{b}^k$ as,

$\begin{array}{c}\frac{1}{6}{\left(3x^2+1\right)}^6={\displaystyle \sum _{k=0}^6\left(\begin{array}{c}6\\ k\end{array}\right)}{\left(3x^2\right)}^{6-k}{\left(1\right)}^k\\ =\frac{1}{6}\left(729x^{12}+1458x^{10}+1215x^8+540x^6+135x^4+18x^2+1\right)\\ =\frac{243x^{12}}{2}+\frac{405x^8}{2}+\frac{45x^4}{2}+243x^{10}+90x^6+3x^2+\frac{1}{6}\mathrm{...}\left(2\right)\end{array}$

From (1) and (2), it is observed that ${\displaystyle \int {\left(3x^2+1\right)}^{5}dx\ne \frac{1}{6}{\left(3x^2+1\right)}^6}+C$

Therefore, the equation “${\displaystyle \int {\left(3x^2+1\right)}^{5}dx=\frac{1}{6}{\left(3x^2+1\right)}^6}+C$” is $\underset{\_}{\text{false}}$.