Question

Evaluate the Integral\(\int {tn(2x + 1)dx} \)

Short answer

A mathematical notion is known as an integral show how infinitesimal data can result in displacement, area, volume, and other concepts. The process of locating integrals is referred to as "integration."

To solve the integral, we should use the parts integration approach.

Step-by-step solution

- Given data

Let, \(l = \int {tn(2x + 1)dx \to 1} \)

Let,\(t = {\rm{ }}2x + 1\)

\(\begin{array}{l}\frac{{dt}}{{dx}} = 2\\dx = \frac{{dt}}{2}\end{array}\)

Substitute above in equation 1

\(\begin{array}{l}i = \int {lm + \left( {\frac{{dt}}{2}} \right)} \\i = \frac{1}{2}\int {\ln + dt \to 2} \end{array}\)

- Using integration by parts

Let's say u = lnt and dv = dt. \(du = \frac{1}{t}dt \to 2\) v=t

Substitute above equation in 2

\(\begin{array}{l}l = \frac{1}{2}\left( {t\ln t - \int {dt} } \right)\\l = \frac{1}{2}(t\ln t - t) + c \to 3\end{array}\)

Where c \( \to \)constant

Rearranging values

By putting t= 2x+1 back in 3

\(l = \frac{1}{2}\left( {(2x + 1)\ln (2x + 1) - (2x + 1)} \right) + c\)

Therefore, the integral value of\(\int {\ln (2x + 1)dx} \)is :

\(\frac{1}{2}((2x + 1)\ln (2x + 1) - (2x + 1)) + c\)