2008 H2 Mathematics Paper 1 Question 5
Integration Techniques
Answers
(n+1)2nen+1+1 Full solutions
(i)
∫0311+9x21dx=91∫03191+x21dx=91[311tan−1(31x)]031=[31tan−13x]031=9π■ (ii)
∫1exnlnxdx=[n+1xn+1lnx]1e−∫1en+1xn+1x1dx=n+1en+1−n+11∫1exndx=n+1en+1−n+11[n+1xn+1]1e=n+1en+1−(n+1)2en+1+(n+1)21=(n+1)2(n+1)en+1−en+1+1=(n+1)2nen+1+1■