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
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