Mathematics

Integrate by using the substitution suggested in bracket $\int {12\left( {{y^4} + 4{y^2} + 1} \right)\left( {{y^3} + 2y} \right)dy,}$ , $\left( {use,u = {y^4} + 4{y^2} + 1} \right)$

SOLUTION
$u=y^{4}+4{y^{2}}+1\implies d{u}=(4{y^{3}}+8{y})d{y}=4(y^{3}+2{y})d{y}$
$\displaystyle\int 12(y^{4}+4{y^{2}}+1)(y^{3}+2{y})d{y}=3\int u d{u}=\dfrac{3}{2}u^{2}+c=\dfrac{3}{2}(y^{4}+2{y^{2}}+1)^{2}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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