Let kx-x2-1-x3-3x-6dx and k-1-1-2- then the value of k-2 is

The correct option is C 2 I = ∫(x^2 + 1)√(x^3 + 3x + 6) dx Let x^3 + 3x + 6 = t dt = 3(x^2 + 1) dx dx = dt3(x^2 + 1) I = ∫dt (x^2 ...

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Byju's Answer

Standard XII

Mathematics

Change of Variables

Let kx=∫x2+...

Question

Let $k (x) = \int \frac{(x^{2} + 1)}{\sqrt[3]{x^{3} + 3 x + 6}} d x$ and $k (- 1) = \frac{1}{\sqrt[3]{2}}$ , then the value of $k (- 2)$ is

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Solution

The correct option is C $2$
$I = \int \frac{(x^{2} + 1)}{\sqrt[3]{x^{3} + 3 x + 6}} d x$

Let $x^{3} + 3 x + 6 = t$

$d t = 3 (x^{2} + 1) d x$

$d x = \frac{d t}{3 (x^{2} + 1)}$

$I = \int \frac{d t (x^{2} + 1)}{3 (x^{2} + 1) t^{1 / 3}}$

$\Rightarrow \int \frac{d t}{3 t^{1 / 3}} = \frac{t^{2 / 3}}{2} + c$

$I = (x^{3} + 3 x + 6)^{2 / 3} + c$

$k (x) = (x^{3} + 3 x + 6)^{2 / 3} + c$

$k (- 1) = \frac{1}{2^{1 / 3}}$

$∴ c = 0$

$k (- 2) = \frac{(- 8 - 6 + 6)^{2 / 3}}{2} + 0$

$= \frac{(- 2)^{2 / 3}}{2} + 0$

$= 2$

$k (- 2) = 2$

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Let k(x)=∫(x2+1)3√x3+3x+6dx and k(−1)=13√2, then the value of k(−2) is

The correct option is C 2I=∫(x2+1)3√x3+3x+6dxLet x3+3x+6=tdt=3(x2+1)dxdx=dt3(x2+1)I=∫dt(x2+1)3(x2+1)t1/3⇒∫dt3t1/3=t2/32+cI=(x3+3x+6)2/3+ck(x)=(x3+3x+6)2/3+ck(−1)=121/3 ∴c=0k(−2)=(−8−6+6)2/32+0=(−2)2/32+0=2k(−2)=2

Let $k (x) = \int \frac{(x^{2} + 1)}{\sqrt[3]{x^{3} + 3 x + 6}} d x$ and $k (- 1) = \frac{1}{\sqrt[3]{2}}$ , then the value of $k (- 2)$ is