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Question 61
Use substitution to find the integral ∫ex(e2x+1) (ex−3) dx. \text { Use substitution to find the integral } \int \frac { e ^ { x } } { \left( e ^ { 2 x } + 1 \right) \left( e ^ { x } - 3 \right) } d x \text {. } Use substitution to find the integral ∫(e2x+1) (ex−3) exdx.
A) ∫ex(e2x+3) (ex−3) dx=124(arctan(ex) ) +C\int \frac { e ^ { x } } { \left( e ^ { 2 x } + 3 \right) \left( e ^ { x } - 3 \right) } d x = \frac { 1 } { 24 } \left( \arctan \left( e ^ { x } \right) \right) + C∫(e2x+3) (ex−3) exdx=241(arctan(ex) ) +C B) ∫ex(e2x+3) (ex−3) dx=124(ln∣e2x+3∣−2arctan(ex) ) +C\int \frac { e ^ { x } } { \left( e ^ { 2 x } + 3 \right) \left( e ^ { x } - 3 \right) } d x = \frac { 1 } { 24 } \left( \ln \left| e ^ { 2 x } + 3 \right| - 2 \arctan \left( e ^ { x } \right) \right) + C∫(e2x+3) (ex−3) exdx=241(lne2x+3−2arctan(ex) ) +C C) ∫ex(e2x+3) (ex−3) dx=124(2ln∣ex−3∣−ln∣e2x+6∣−4arctan(e2x) ) +C\int \frac { e ^ { x } } { \left( e ^ { 2 x } + 3 \right) \left( e ^ { x } - 3 \right) } d x = \frac { 1 } { 24 } \left( 2 \ln \left| e ^ { x } - 3 \right| - \ln \left| e ^ { 2 x } + 6 \right| - 4 \arctan \left( e ^ { 2 x } \right) \right) + C∫(e2x+3) (ex−3) exdx=241(2ln∣ex−3∣−lne2x+6−4arctan(e2x) ) +C D) ∫ex(e2x+3) (ex−3) dx=120(2ln∣ex−3∣−ln∣e2x+1∣−6arctan(ex) ) +C\int \frac { e ^ { x } } { \left( e ^ { 2 x } + 3 \right) \left( e ^ { x } - 3 \right) } d x = \frac { 1 } { 20 } \left( 2 \ln \left| e ^ { x } - 3 \right| - \ln \left| e ^ { 2 x } + 1 \right| - 6 \arctan \left( e ^ { x } \right) \right) + C∫(e2x+3) (ex−3) exdx=201(2ln∣ex−3∣−lne2x+1−6arctan(ex) ) +C E) ∫ex(e2x+3) (ex−3) dx=124(2ln∣e2x−6∣−ln∣e2x+6∣−2arctan(ex) ) +C\int \frac { e ^ { x } } { \left( e ^ { 2 x } + 3 \right) \left( e ^ { x } - 3 \right) } d x = \frac { 1 } { 24 } \left( 2 \ln \left| e ^ { 2 x } - 6 \right| - \ln \left| e ^ { 2 x } + 6 \right| - 2 \arctan \left( e ^ { x } \right) \right) + C∫(e2x+3) (ex−3) exdx=241(2lne2x−6−lne2x+6−2arctan(ex) ) +C
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Q57: Find the indefinite integral.
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