Question

\[ \begin{array}{l} \int_{0}^{2} \frac{x}{2 x^{2}+1} d x \\ \int_{1}^{2}\left(\frac{2}{x^{2}}-3\right) d x \\ \int_{1}^{4}\left(2^{x+1}-3\right) d x \\ \int^{6} x+2 \end{array} \]

          \[
\begin{array}{l}
\int_{0}^{2} \frac{x}{2 x^{2}+1} d x \\
\int_{1}^{2}\left(\frac{2}{x^{2}}-3\right) d x \\
\int_{1}^{4}\left(2^{x+1}-3\right) d x \\
\int^{6} x+2
\end{array}
\]

∫0^2(x)/(2 x^2+1) d x
∫1^2((2)/(x^2)-3) d x
∫1^4(2^x+1-3) d x
∫^6 x+2

Added by John D.

Engineering Mathematics

Stroud, K. A. Booth, Dexter J. 7th Edition

Chapter 8

Thanks for your feedback!

\[ \begin{array}{l} \int_{0}^{2} \frac{x}{2 x^{2}+1} d x \\ \int_{1}^{2}\left(\frac{2}{x^{2}}-3\right) d x \\ \int_{1}^{4}\left(2^{x+1}-3\right) d x \\ \int^{6} x+2 \end{array} \]