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Exercise 11.2 · Q1

Q.Integrate the following functions with respect to xx:

(i) (x+5)6(x+5)^{6}
(ii) 1(2−3x)4\dfrac{1}{(2-3x)^{4}}
(iii) 3x+2\sqrt{3x+2}
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Each integrand is a power of a linear argument ax+bax+b; apply the power rule to the argument as if it were xx, then divide by the coefficient aa.

Step 1. Part (i). Here a=1, n=6a=1,\ n=6, so no correction factor is needed beyond dividing by a=1a=1.

∫(x+5)6 dx=(x+5)77+c.\int (x+5)^6\,dx=\frac{(x+5)^7}{7}+c.

Step 2. Part (ii). Rewrite as (2−3x)−4(2-3x)^{-4}; here a=−3, n=−4a=-3,\ n=-4.

∫(2−3x)−4 dx=1−3⋅(2−3x)−3−3+c=(2−3x)−39+c=19(2−3x)3+c.\int (2-3x)^{-4}\,dx=\frac{1}{-3}\cdot\frac{(2-3x)^{-3}}{-3}+c=\frac{(2-3x)^{-3}}{9}+c=\frac{1}{9(2-3x)^3}+c.

Step 3. Part (iii). Rewrite as (3x+2)1/2(3x+2)^{1/2}; here a=3, n=12a=3,\ n=\tfrac12.

∫(3x+2)1/2 dx=13⋅(3x+2)3/23/2+c=13⋅23(3x+2)3/2+c=29(3x+2)3/2+c.\int (3x+2)^{1/2}\,dx=\frac13\cdot\frac{(3x+2)^{3/2}}{3/2}+c=\frac13\cdot\frac23(3x+2)^{3/2}+c=\frac29(3x+2)^{3/2}+c.

Step 4. Check by differentiating. ddx ⁣((x+5)77)=(x+5)6\dfrac{d}{dx}\!\left(\dfrac{(x+5)^7}{7}\right)=(x+5)^6 ✓. ddx ⁣(19(2−3x)3)=19⋅(−3)(2−3x)−4⋅(−3)=99(2−3x)−4=(2−3x)−4\dfrac{d}{dx}\!\left(\dfrac{1}{9(2-3x)^3}\right)=\dfrac19\cdot(-3)(2-3x)^{-4}\cdot(-3)=\dfrac{9}{9}(2-3x)^{-4}=(2-3x)^{-4} ✓. ddx ⁣(29(3x+2)3/2)=29⋅32(3x+2)1/2⋅3=(3x+2)1/2\dfrac{d}{dx}\!\left(\dfrac29(3x+2)^{3/2}\right)=\dfrac29\cdot\dfrac32(3x+2)^{1/2}\cdot3=(3x+2)^{1/2} ✓.

✓Final answer

(i) (x+5)77+c\dfrac{(x+5)^{7}}{7}+c (ii) 19(2−3x)3+c\dfrac{1}{9(2-3x)^{3}}+c (iii) 29(3x+2)3/2+c\dfrac29(3x+2)^{3/2}+c

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