Mathematics

# Solve $\int {x.\sqrt[5]{{x + 1}}dx}$

##### SOLUTION
$\int x.\sqrt[5]{x+1}dx...(1)$
put $x+1=t^{5}….(2)$
hence, $x=t^{5}- 1...(3)$  and $dx=5t^{4} dt...(4)$
substitute equation (2) (3) and (4) in equation (1) we get
$\int(t^{5}-1)(t)(5t^{4})dt$
$\int5(t^{5}-1)t^{5}dt$
$5\int(t^{10}-t^{5})dt$
$5\bigg[\dfrac{t^{11}}{11}-\dfrac{t^{6}}{6}\bigg]$
now, put value of $t=(x+1)^{\dfrac{1}{5}}$ from equation (2) in above equation we get.
$I=5\bigg[ \dfrac{(x+1)^{\dfrac{11}{5}}}{11}-\dfrac{(x+1)^{\dfrac{6}{5}}}{6}\bigg]+C$

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

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