Answer to Question #250544 in Calculus for JaytheCreator

Question

Find (i) ∫ 𝑑𝑥 𝑥(ln 𝑥)2

(ii) ∫ 𝑥√𝑥 + 1 𝑑𝑥

(iii) ∫ 𝑥(𝑥 2 + 3) 4𝑑𝑥 1 0

(iv) cos2 𝑥 sin3 𝑥 𝑑𝑥. Try 𝑡 = sin 𝑥 

Answer 1

Solution:

(i) $I=\int \dfrac{dx}{x (\ln x)^2}$ ...(i)

Put $\ln x=t$

$\Rightarrow\dfrac 1xdx=dt$

So, (i) becomes,

$I=\int \dfrac{dt}{ (t)^2} \\=\int t^{-2}dt \\=-t^{-1}+C \\=-\dfrac 1t+C \\=-\dfrac 1{\ln x}+C$

(ii) $I=\int \dfrac{x}{\sqrt{x+1}}dx$ ...(ii)

Put $x+1=t$

$\Rightarrow dx=dt$

So, (ii) becomes,

$I=\int \dfrac{t-1}{\sqrt t}dt \\=\int (\dfrac{t}{\sqrt t}-\dfrac{1}{\sqrt t})dt \\=\int( t^{1/2}-t^{-1/2})dt \\=\dfrac{t^{3/2}}{3/2}-\dfrac{t^{1/2}}{1/2}+C \\=\dfrac 23(x+1)^{3/2}-2(x+1)^{1/2}+C$

(iii) $I=\int \dfrac {x}{(x^2+3)^4}dx$ ...(iii)

Put $x^2+3=t$

$\Rightarrow 2xdx=dt \\\Rightarrow xdx=\dfrac{dt}2$

So, (iii) becomes,

$I=\dfrac 12\int \dfrac {dt}{(t)^4} \\=\dfrac 12\int t^{-4}{dt} \\=\dfrac 12(\dfrac{t^{-3}}{-3})+C \\=-\dfrac 1{6(x^2+3)^3}+C$

(iv) $I=\int \cos2x\sin 3x\ dx$

$=\dfrac12\int (2 \cos2x\sin 3x)dx \\=\dfrac12\int (\sin(2x+3x)-\sin(2x-3x))dx \\=\dfrac12\int (\sin(5x)-\sin(-x))dx \\=\dfrac12\int (\sin(5x)+\sin(x))dx \\=\dfrac12 (-\dfrac{\cos(5x)}5-\cos x)+C$