Question

integration of logx in limit of 1 to 2​

Answer 1

Answer:

2log2 - 1

Step-by-step explanation:

By using byparts method ILATE

$\int\limits^2_1 {logx} \, dx$

=$\int\limits^2_1 {1 . logx} \, dx \\= \limits^2_1 [logx\int\limits {1} \, dx] -\int\limits^2_1 {\frac{d}{dx} logx [\int\limits {1} \, dx ]} \, dx \\=\limits^2_1 [xlogx] - \int\limits^2_1 {\frac{1}{x}(x) } \, dx \\=[2log2-1log1] - \int\limits^2_1 {1} \, dx \\=2log2 - \limits^2_1[x]\\=2log2-[2-1]\\=2log2-1$        Since log1 = 0