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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.2
Integrate the functions in Exercises 23 to 59.
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Question : 34 of 271
Marks: +1, -0
(x3−1)1/3x5(x^3-1)^{1/3}x^5
Solution:  
Let I=∫(x3−1)1/3x5dxI=\int (x^3-1)^{1/3}x^5dx
=13∫(x3−1)1/3x3⋅3x2dx=\frac{1}{3}\int (x^3-1)^{1/3}x^3\cdot 3x^2dx
Put x3−1=tx^3 - 1 = t ⇒3x2dx=dt\Rightarrow 3x^2 dx = dt . Also x3=t+1x^3 = t + 1
∴I=13∫t1/3(t+1)dt\therefore I = \frac{1}{3}\int t^{1/3}(t+1)dt =13[37t7/3+34t4/3]+C=\frac{1}{3}\left[ \frac{3}{7}t^{7/3}+\frac{3}{4}t^{4/3} \right]+C
=17t7/3+14t4/3+C=\frac{1}{7}t^{7/3}+\frac{1}{4}t^{4/3}+C
=17(x3−1)7/3+14(x3−1)4/3+C=\frac{1}{7}(x^3-1)^{7/3}+\frac{1}{4}(x^3-1)^{4/3}+C
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