Integrate each of the following with respect to x :
Question 1 :
Question 2 :
x/√1 + x2
Question 3 :
x2/1 + x6
Question 4 :
(ex - e-x)/(ex + e-x)
Question 5 :
(10x9 + 10x loge10)/(10x + x10)
Question 6 :
(sin √x)/√x
1. Answer :
In the function, the power of 'e' is x3. The derivative is x3 is 3x2 and it is being as a part of the function and we will be able to write the 3x2 along with dx using multiplication.
Now substitute a new variable for x3 for which we have the derivative 3x2 being a part of the given function.
Let u = x3.
u = x3
Differentiate with respect to x.
du/dx = 3x2
Multiply both sides by dx.
du = 3x2dx
Substitute x3 = u and 3x2dx = du in (1)
= ∫eudu
= eu + C
Substitute u = x3.
2. Answer :
= ∫(x/√1 + x2)dx
= ∫(xdx)/√1 + x2 ----(1)
Let t = 1 + x2.
t = 1 + x2
Differentiate with respect to x.
dt/dx = 0 + 2x
dt/dx = 2x
Multiply both sides by dx.
dt = 2xdx
Divide both sides by 2.
dt/2 = xdx
Substitute 1 + x2 = t and xdx = dt/2 in (1).
∫(xdx)/√1 + x2 = ∫(dt/2)/√t
= (1/2)∫dt/√t
= (1/2)∫dt/t-1/2
= (1/2)∫t-1/2dt
= (1/2)t-1/2 + 1/(-1/2 + 1) + C
= (1/2)t1/2/(1/2) + C
= (1/2)t1/2(2/1) + C
= t1/2 + C
= √t + C
Substitute t = x2 + 1.
= √(1 + x2) + C
3. Answer :
= ∫[x2/(1 + x6)] dx
= ∫(x2dx)/[1 + (x3)2] ----(1)
Let t = x3.
t = x3
Differentiate with respect to x.
dt/dx = 3x2
Multiply both sides by dx.
dt = 3x2dx
Divide both sides by 3.
dt/3 = x2dx
Substitute x3 = t and x2dx = dt/3 in (1)
∫(x2dx)/(1 + (x3)2 = ∫(dt/3)/(1 + t2)
= (1/3)∫1/(1 + t2)dt
= (1/3)tan-1t + c
Substitute t = x3.
= (1/3)tan-1(x3) + c
4. Answer :
= ∫[(ex - e-x)/(ex + e-x)]dx
= ∫(ex - e-x)dx/(ex + e-x) ----(1)
Let t = ex + e-x.
t = ex + e-x
Differentiate with respect to x.
dt/dx = ex + e-x(-1)
dt/dx = ex - e-x
Multiply both sides by dx.
dt = (ex - e-x)dx
Substitute ex + e-x = t and (ex - e-x)dx = dt in (1).
∫(ex - e-x)dx/(ex + e-x) = ∫dt/t
= ∫(1/t)dt
= lnt
Substitute t = ex + e-x.
= ln(ex + e-x) + c
5. Answer :
= ∫[(10x9 + 10xln10)/(10x + x10)]dx
= ∫(10x9 + 10xln10)dx/(10x + x10) ----(1)
Let u = 10x + x10.
u = 10x + x10
Differentiate with respect to x.
du/dx = 10xln10 + 10x9
Multiply both sides by dx.
du = (10xln10 + 10x9)dx
du = (10x9 + 10xln10)dx
Substitute 10x + x10 = u and (10x9 + 10xln10)dx in (1).
= ∫du/u
= ∫(1/u)du
= lnu + c
Substitute u = 10x + x10.
= ln(10x + x10) + c
6. Answer :
= ∫[(sin√x)/√x]dx
= ∫(sin√x)(dx/√x) ----(1)
Let u = √x.
u = √x
u = x1/2
du/dx = (1/2)x1/2 - 1
du/dx = (1/2)x-1/2
du/dx = 1/(2x1/2)
du/dx = 1/(2√x)
Multiply both sides by 2dx.
2du = dx/√x
Substitute √x = u and dx/√x = 2du in (1).
∫(sin√x)(dx/√x) = ∫(sinu)(2du)
= 2∫sinudu
= 2(-cosu) + c
= -2cosu + c
Substitute u = √x.
= -2cos√x + c
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