Question

∫(5dx/1x^2), ∫(x^3-3x+sinx)dx, ∫(2x+1)^4dx, ∫sin3xdx

Answer 1

5∫x⁻²dx=-5*x⁻¹+c

∫(x^3-3x+sinx)dx=x⁴/4-3x²/2-cosx+c

 ∫(2x+1)^4dx,   2x+1=z   2dx=dz    dx=dz/2
 1/2∫z⁴dz=1/(2*5)z⁵=1/10 z⁵+c
 ∫ sin3xdx   3x=z  dx=dz/3
 ∫sin3xdx = 1/3∫sinzdz=-cosz/3=-cos3x/3

Answer 2

Let's solve each integral step by step:

1. ∫(5dx/1x^2):

To solve this integral, we can rewrite it as: ∫5x^(-2)dx

Now, applying the power rule of integration, we increase the exponent by 1 and divide by the new exponent: ∫5x^(-2)dx = (5/(-1 + 1))x^(-1) + C = -5x^(-1) + C = -5/x + C

Therefore, ∫(5dx/1x^2) = -5/x + C.

2. ∫(x^3 - 3x + sin(x))dx:

To integrate this expression, we can use the power rule of integration for each term. Applying the power rule, we have: ∫x^3dx = (1/4)x^4 + C ∫-3xdx = -(3/2)x^2 + C ∫sin(x)dx = -cos(x) + C

Combining these results, we get: ∫(x^3 - 3x + sin(x))dx = (1/4)x^4 - (3/2)x^2 - cos(x) + C

Therefore, ∫(x^3 - 3x + sin(x))dx = (1/4)x^4 - (3/2)x^2 - cos(x) + C.

3. ∫(2x + 1)^4dx:

To integrate this expression, we can expand the binomial using the binomial theorem, and then apply the power rule of integration term by term. The binomial expansion of (2x + 1)^4 is: (2x + 1)^4 = 16x^4 + 32x^3 + 24x^2 + 8x + 1

Now, we integrate each term using the power rule: ∫16x^4dx = (16/5)x^5 + C ∫32x^3dx = (8/1)x^4 + C ∫24x^2dx = (8/1)x^3 + C ∫8xdx = (4/1)x^2 + C ∫1dx = x + C

Combining these results, we get: ∫(2x + 1)^4dx = (16/5)x^5 + (8/1)x^4 + (8/1)x^3 + (4/1)x^2 + x + C

Therefore, ∫(2x + 1)^4dx = (16/5)x^5 + 8x^4 + 8x^3 + 4x^2 + x + C.

4. ∫sin(3x)dx:

To integrate sin(3x), we can use the trigonometric identity: ∫sin(ax)dx = -(1/a)cos(ax) + C

Applying this identity to our integral, where a = 3, we have: ∫sin(3x)dx = -(1/3)cos(3x) + C

Therefore, ∫sin(3x)dx = -(1/3)cos(3x) + C.

Please note that "C" represents the constant of integration, and it is added to account for the family of antiderivatives that differ by a constant.