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प्रश्न
Evaluate: `int log ("x"^2 + "x")` dx
उत्तर
Let I = `int log ("x"^2 + "x")` dx
`= int log ("x"^2 + "x") * 1 * "dx"`
`= log ("x"^2 + "x") int 1 * "dx" - int {"d"/"dx" log ("x"^2 + "x") int 1 * "dx"}`dx
`= log ("x"^2 + "x") * "x" - int 1/("x"^2 + "x") * (2"x" + 1) * "x" * "dx"`
`= "x" * log ("x"^2 + "x") - int 1/("x"("x + 1")) * (2"x" + 1) * "x" * "dx"`
`= "x" * log ("x"^2 + "x") - int("2x + 1")/("x + 1")`dx
`= "x" * log ("x"^2 + "x") - int((2"x" + 2) - 1)/("x + 1")` dx
`= "x" * log ("x"^2 + "x") - int[(2("x + 1"))/("x + 1") - 1/("x + 1")]` dx
`= "x" * log ("x"^2 + "x") - int[2 - 1/"x + 1"]` dx
`= "x" * [log("x"^2 + "x")] - (2"x" - log |"x + 1"|) + "c"`
∴ I = `"x" * [log("x"^2 + "x")] - 2"x" + log |"x + 1"|` + c
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