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Let u = tx,du = x dtL =∫(0~1)ƒ;[tx] dt= [1/x]∫(0~x)ƒ;[u] du = ƒ;[x] + xsinx∫(0~x)ƒ;[u] du = xƒ;[x] + x²;sinxƒ;[x] = xƒ;'[x] + ƒ;[x] + 2xsinx + x²;cosx…
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