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Given that y > 0, find ∫((3y - 4)/y(3y + 2)) dy (taken from the Edexcel C4 2016 paper)

This can't be integrated directly as y appears in the numerator and denominator. This is an indication that you must integrate by parts. A/y + B/(3y+2) = (3y - 4)/y(3y + 2) A and B must be found. Multiply by...
SS
Answered by Saskia S. • Maths tutor
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The complex conjugate of 2-3i is also a root of z^3+pz^2+qz-13p=0. Find a quadratic factor of z^3+pz^2+qz-13p=0 with real coefficients and thus find the real root of the equation.

z-2+3i times z-2-3i = z 2 -4z+13. z 3 -2z 2 +5z+26 divided by z 2 -4z+13 = z+2. Therefore the real root is z=-2.
WN
Answered by William N. • Maths tutor
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Given that 2-3i is a root to the equation z^3+pz^2+qz-13p=0, show that p=-2 and q=5.

Substitute 2-3i into equation using part i (2-3i) 3 =-46-9i. -46-9i+p(-5-12i)+q(2-3i)-13p=0. -46-18p+2q-9i-12pi-3iq=0. Real: -46-18p+2q=0 and Imaginary: -9-12p-3q=0. p=-2, q=5
WN
Answered by William N. • Maths tutor
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Show (2-3i)^3 can be expressed in the form a+bi where a and b are negative integers.

(2-3i) x (2-3i) = -5-12i. -5-12i x (2-3i) = -46-9i. a=-46, b=-9
WN
Answered by William N. • Maths tutor
4447 Views

Integrate by parts the following function: ln(x)/x^3

Let integrate be denoted by the letter I. For instance I(f) is the integration of a function f . Then Integration by parts states that I(u v') = uv - I(u' v), where u,v are function with u', v' their respect...
PD
Answered by Paul D. • Maths tutor
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