1 x² + 4x + 8 dx 0³ 1 + sin 04 de ex 1 - tan ex dx 1 √ [1 + (In 1)²] dt

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answerhappygod
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1 x² + 4x + 8 dx 0³ 1 + sin 04 de ex 1 - tan ex dx 1 √ [1 + (In 1)²] dt

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1 X 4x 8 Dx 0 1 Sin 04 De Ex 1 Tan Ex Dx 1 1 In 1 Dt 1
1 X 4x 8 Dx 0 1 Sin 04 De Ex 1 Tan Ex Dx 1 1 In 1 Dt 1 (55.47 KiB) Viewed 26 times
Solve the indefinite integrals.
1 x² + 4x + 8 dx 0³ 1 + sin 04 de ex 1 - tan ex dx 1 √ [1 + (In 1)²] dt
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