(i) – cosx 2 + C (ii)  n |x 2 + 1| + C

(iii) (tanx) 2 + C or + C (iv)  n |e x + x| + C

(v)  n |x + cosx| + C (vi)  n |e 2x – 2| + C

(vii)  n |x 2 + sin2x + 2x| + C (viii)  n |  n(secx + tanx) | + C

(ix) (x + 2) 3/2 – 4(x + 2) 1/2 + C (x) (e 2x – e – 2x ) + 2x + C

(xi) e 3x + e 2x + e x + C (xii) –  n + C

(xiii) – + C (xiv) + C

Sol. (i) Ι = Put x 2 = t ⇒ x dx =

⇒ Ι = = – + C = – + C

(ii) Put x 2 + 1 = t ⇒ x dx =

⇒ Ι = =  n(x 2 + 1) + C

(iii) Ι = Put tanx = t ⇒ sec 2 x dx = dt

⇒ Ι = = + C

(iv) Ι = Put e x + x = t ⇒ (e x + 1) dx = dt

⇒ Ι =  n(e x + x) + C

(v) Let x + cosx = t ⇒ (1– sinx) dx = dt

=  n |x + cosx| + C

(vi) where e 2x – 2 = t ⇒ 2e 2x dx = dt

=  n |t| + C =  n|e 2x – 2| + C

(vii) Let x 2 + sin2x + 2x = t

2(x + cos2x + 1) dx = dt

=  n|x 2 + sin2x + 2x| + C

(viii) Let  n(secx + tanx) = t ⇒ secx dx = dt

∴ =  nt + C =  n |  n(secx + tanx)| + C

(ix)

= –

= – 2 × 2 + C

= (x + 2) 3/2 – 4(x + 2) 1/2 + C

(x) Let e x = t ⇒ e x dx = dt dx =

Ι = = = =

(xi) Put t = e x ⇒ dt = e x dx

= = = =

(xii) Ι = Let x 5 = t ⇒ 5x 4 dx = dt

Ι = =

= [  n|t| –  n |(t + 1)|] + C

= ln + C = –  n + C

(xiii) Ι =

Ι =

Let = t ⇒

Ι =

= =

(xiv) Ι =

= Put 1 – = t 2 ⇒ = dt

⇒ Ι =

= × = = + C

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