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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ dx x + x 3 = a x + b(x 3) + c(x 6) + dln ( x6 + 1 ) + e, e...

If ∫dxx+x3=ax+b(x3)+c(x6)+∫ \dfrac{dx}{\sqrt{x} + x 3} = a\sqrt{x} + b\left(x 3\right) + c\left(x 6\right) +dln(x6+1)+e,edln ( x6 + 1 ) + e, e being an arbitrary constant, then the value of 20a+b+c+d20a + b + c + d is

Correct answer

37

Step-by-step solution

Let x=u6,dx=6u5dux = {u}^{6}, dx = 6{u}^{5}du
∫dxx+x3=∫6u5duu3+u2=6∫u3u+1du∫ \dfrac{dx}{\sqrt{x} + x3} = ∫ \dfrac{6{u}^{5}du}{{u}^{3} + {u}^{2}} = 6 ∫ \dfrac{{u}^{3}}{u + 1}du
=6∫(u2−u+1−1u+1)du= 6 ∫ \left({u}^{2} - u + 1 - \dfrac{1}{u + 1}\right)du
=2u3−3u2+6u−6ln⁡(u+1)+e= 2{u}^{3} - 3{u}^{2} + 6u - 6\ln \left(u + 1\right) +e
=2x−3(x3)+6(x6)−6ln⁡(x6+1)+e= 2\sqrt{x} - 3\left(x3\right) + 6\left(x6\right) - 6\ln \left(x6 + 1\right) +e
∴a=2,b=−3,c=6,d=−6∴ a = 2, b = - 3, c = 6, d = - 6
∴20a+b+c+d=37∴ 20a + b + c + d = 37
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