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QUESTION: 1

Evaluate:

Solution:

QUESTION: 2

The value of is :

Solution:

QUESTION: 3

Evaluate:

Solution:

QUESTION: 4

Evaluate:

Solution:

∫dx/(x^{2} - a^{4}b^{4})

= ∫dx/(x^{2} - a^{2}b^{2})^{2}

= ∫dx/(x^{2} - a^{2}b^{2})(x^{2} + a^{2}b^{2})

As by formula dx/(n - n)(n + n) = 1/2n log|(x - n)/(x + n)|

= 1/2n log|(x - a^{2}b^{2})/(x + a^{2}b^{2})|

QUESTION: 5

Simplify the integrand of

Solution:

Let I = ∫5x dx/[(x+1)(x2+9)]

Let 5x/[(x+1)(x2+9)] = A(x+1) + (Bx + C)/(x^{2}+9)

⇒5x = (A+B)x^{2} + (B+C)x + 9A + C

Comparing the coefficients of x2 on both sides, we get

A + B = 0 ...(1)

Comparing the coefficients of x on both sides, we get

B + C = 5 .....(2)

Comparing constants on both sides, we get

9A +C = 0 ....(3)

From (1), we get B = −A

From (3), we get C = −9A

now, from (2), we get

−A − 9A = 5

⇒A = −1/2

B = 1/2

C = 9/2

So, 5x/[(x+1)(x^{2}+9)] = −1/2 × [1(x+1)] + 1/2 × (x + 9)/(x^{2 }+ 9)

So, ∫5x dx/[(x+1)(x^{2 }+ 9)] = −1/2∫dx/(x+1) + 1/2∫(x+9)/(x^{2 }+9) dx

=−1/2 log|x+1| + 1/2∫x dx/(x^{2} +9) + 9/2∫dx (x^{2 }+9)

=−1/2 log|x+1| + 1/4∫2x dx/(x^{2 }+9) + 9/2∫dx/(x^{2 }+(3)^{2})

⇒−1/2 log |x+1| + 1/4log|x ^{2} +9| + 9/2×1/3 tan^{−1}(x/3) + C

=−1/2 log |x+1| + 1/4log∣|x2+9| + 3/2 tan^{−1}(x/3) + C

QUESTION: 6

Evaluate:

Solution:

∫dx/x(x^{n} + 1)..............(1)

∫dx/x(x^{n} + 1) *(x^{n} - 1)/(x^{n} - 1)

Put x^{n} = t

dt = nx^{(n-1)}dx

dt/n = x^{(n-1)}dx

Put the value of dt/n in eq(1)

= ∫(1/n)dt/t(t+1)

= 1/n ∫dt/(t+1)t

= 1/n{∫dt/t - ∫dt/t+1}

= 1/n {ln t - lnt + 1} + c

= 1/n {ln |t/(t + 1)|} + c

= 1/n {ln |x^{n}/(x^{n} + 1)|} + c

QUESTION: 7

How would you split the rational function f(x) into partial function?

Solution:

QUESTION: 8

The integral of is:

Solution:

∫dx/x^{3}(x^{-2} -4).............(1)

= ∫x^{-3} dx/(x^{-2} - 4)

Let t = (x^{-2} - 4)

dt = -2x^{-3} dx

x^{-3} = -dt/2

Put the value of x^{-3} in eq(1)

= -½ ∫dt/t

= -½ log t + c

= -½ log(x^{-2} - 4) + c

= -½ log(1-4x^{2})/x^{2} + c

= ½ log(x^{2}/(1 - 4x^{2})) + c

QUESTION: 9

The value of

Solution:

∫1/cos^{2} x(1 - tanx)(2 - tanx) dx

= ∫(sec^{2} x)/(2 - 3tanx + tan^{2}x) dx

Put t = tanx

dt = sec^{2}x dx

∫dt/(2 - 3t + t^{2})dx

∫dt/(t^{2} - 3t + 9/4) + (2 - 9/4)

= ∫dt/((t - 3/2)^{2} - (½)

Using the formula ∫dx/(x^{2} - a^{2}), we get

= ½[½] log |[t - 3/2 - ½]/[t - 3/2 + ½]| + c

= log|(t - 2)/(t - 1)| + c

Taking - common, we get

= log|(2 - t)/(1 - t)| + c

log|(2 - tanx)/(1 - tanx)| + c

QUESTION: 10

Correct evaluation of

Solution:

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