<p>Use the substitution $u={{e}^{x}}$ to show that $\int_{0}^{1}{\frac{{{\mathrm{e}}^{x}}-1}{{{\mathrm{e}}^{x}}+1}}\mathrm{ d}x=\int_{1}^{\mathrm{e}}{\frac{u-1}{u\left( u+1 \right)}}\mathrm{ d}u$. Hence, by using partial fractions, find the exact value of $\int_{0}^{1}{\frac{{{\mathrm{e}}^{x}}-1}{{{\mathrm{e}}^{x}}+1}}\mathrm{ d}x$. Suggest an alternative method.</p>

Use the substitution $u={{e}^{x}}$ to show that $\displaystyle\displaystyle \int_{0}^{1}{\frac{{{\mathrm{e}}^{x}}-1}{{{\mathrm{e}}^{x}}+1}}\mathrm{ d}x=\displaystyle\displaystyle \int_{1}^{\mathrm{e}}{\frac{u-1}{u\left( u+1 \right)}}\mathrm{ d}u$. Hence, by using partial fractions, find the exact value of $\displaystyle\displaystyle \int_{0}^{1}{\frac{{{\mathrm{e}}^{x}}-1}{{{\mathrm{e}}^{x}}+1}}\mathrm{ d}x$. Suggest an alternative method.

Techniques of  Integration - General

Techniques of  Integration

H2 Math

International Baccalaureate Higher Level Mathematics

N1995 P1 Q16 [Modified]

N1995 P1 Q16 [Modified]

Video Solution: