19/08/16 · I = ∫ln(1 x)dx We will use integration by parts, which takes the form ∫udv = uv − ∫vdu So, for ∫ln(1 x)dx, let {u = ln(1 x) ⇒ du = 1 1x dx dv = dx ⇒ v = x Fitting this into the integration by parts formula I = xln(1 x) − ∫ x 1 x dx In integrating the second bit, you could long divide, but this is simplerPlease enter a title Please enter a message Your discussion will live here (Start typing, we will pick a forum for you) Please select a forum Change forum View more forums View less forums GCSEs Alevels/12/19 · 𝑑𝑥 Let 𝑥log𝑥= 𝑡 Differentiating both sides 𝑤𝑟𝑡𝑥 11/𝑥= 𝑑𝑡/𝑑𝑥 (𝑥 1)/𝑥= 𝑑𝑡/𝑑𝑥 " " 𝑑𝑥 = ((𝑥 )/(𝑥 1))𝑑𝑡 Now, our function becomes ∫1 〖" " ((𝑥 1) (𝑥 log𝑥 )^2)/𝑥〗 𝑑𝑥 Putting the value of 𝑥−𝑙𝑜𝑔𝑥=𝑡 & 𝑑𝑥=((𝑥 )/(𝑥 1))𝑑𝑡 = ∫1 〖" " ((𝑥 1) (𝑡)^2)/𝑥〗 (𝑥 )/((𝑥 1) ) 𝑑𝑡 = ∫1 〖" " 𝑡^2 E X X 1 Xlogx Mathematics Topperlearning Com 7ufk1877 Integrate ʃ x log(1+x) dx
