Question 1 :- ∫ e x ( 12 x 2+29 x +5 ) dx

Solution :- ∫ e x ( 12 x 2+29 x +5 ) dx

∫ e x ( 12 x 2+5 x +24 x +5 ) dx

∫ e [ ( 12 x +5 x ) + ( 24 x +5 ) ] dx
x 2

TRICK
∫ e x [ f ( x ) + f ' ( x ) ] dx=e x ⋅f ( x ) +C
2
We have f ( x )=12 x +5 x
'
Then f ( x )=24 x +5
Hence the solution is

e x ( 12 x 2 +5 )

Question 2 :- ∫ cos x−( )

−1 x
dx
√ 1−x 2

Solution :- ( )
x
∫ cos−1 x− dx
√ 1−x 2
x
∫ cos −1 xdx− ∫ dx
√ 1−x 2
TRICK

∫ [ f ( x )+ x f ' ( x ) ] dx=xf ( x ) +C
−1
Here have f ( x ) =cos x
' −1
Then f ( x )=
√1−x 2
Hence the solution is
−1
x cos x+ C

Question 3 :- ∫
1
2
dx
x – 15 x +56

Solution :-
1
∫ 2
dx
x −15 x+56
1
∫ dx
( x – 7) ( x – 8)
TRICK
 1
=
1 1
(
−
1
( x – α) ( x – β ) (α – β ) x – α x – β )
1
Here we have f ( x ) =
( x−7 ) ( x – 8 )
Which can be represented as

f ( x )=
1
=
1
(1
−
1
( x – 7 ) ( x – 8 ) ( 8−7 ) x−8 x−7
=
1
−)(
1
x−8 x−7 )
Integrating

∫ ( x 1– 8 − x 1– 7 ) dx
1 1
∫ dx−∫ dx
x–8 x−7
l og ( x−8 ) −log ( x−7 ) + c

(x – 8 )
l og +c
(x – 7 )
Question 4 :- Which one is true statement
π
I 1=∫ cos 2 x dx , 0< x <
4
π
I 2=∫ sin 2 x dx , 0< x<
4
( 1 ) I 1=I 2 (2) I 1> I 2

(3) I 1< I 2 ( 4 ) ¿

Solution :-
sin 2 x
I 1=∫ cos 2 xdx= +C
2
−cos 2 x
I 2=∫ sin 2 xdx = +C
2
 Red line denotes I 1

Blue line denotes I 2

Hence the correct answer is option (2)

Question 5 :- ∫ √ 2 x+3 ( 4 x +5 ) dx

Solution :- ∫ √ 2 x+3 ( 4 x +5 ) dx
TRICK
∫ √ Linear ( Linear ) dx

∫ √ ax +b ( cx+ d ) dx

Assume √ ax+ b=t

2
ax +b=t
2
t −b
x=
a
Now we can replace cx +d by

( )
2
t −b
cx +d =c +d
a

Here √2 x+ 3=t
2
t −3
¿ x=
2

( )
2
t −3
4 x+5 ¿ 4 +5
2
¿ ¿
2t
dx ¿ dt
2
dx ¿ tdt

Now, we can accumulate above results in main equation

∫ √ 2 x +3 ( 4 x+5 ) dx
∫ t2 ( 2 t2 −1 ) tdt
∫ t 3 ( 2 t 2−1 ) dt
¿∫ ( 2 t −t ) dt
5 3

6 4
t t
¿ −
3 4
¿

Question 6 :- ∫ e3 x sin 9 x dx

Solution :- TRICK is
 ax
ax e (
∫ e ⋅ sin bx dx= 2 2
a sin bx−b cos bx ) +c
a +b

In this problem a = 3 and b =9

Hence
3x
3x e
∫ e ⋅sin 9 x dx= 2 2 ( 3 sin 9 x−9 cos 9 x ) +c
3 +9
3x
3x e (
∫ e ⋅sin 9 x dx= 3sin 9 x−9 cos 9 x )+ c
90

Question 7 :- ∫ e3 x cos 9 x dx

Solution :- TRICK is
ax
ax e (
∫ e ⋅ cos b x dx= 2 2
a cos bx +b sin bx ) +c
a +b

In this problem a = 3 and b =9

Hence
3x
3x e (
∫ e ⋅cos 9 x dx= 2 2
3 cos 9 x +9 sin 9 x ) +c
3 +9
3x
3x e (
∫ e ⋅cos 9 x dx= 3 cos 9 x +9 sin 9 x )+ c
90

Question 8 :-
dx
∫ ( 17 )
x x +1

Solution :-
dx
∫ ( 17 )
x x +1

dx
∫ n∈ N,
x ( x n +1 )

Note :- the value of “ n “ must be NATURAL number to use the below concept.

Take x n common & put 1+ x−n=t

Take x 17 common

dx
∫
x
18
( 1+
x
1
17 )
dx
∫
x ( 1+ x−17 )
18
 −17
Put 1+ x =t
−18
−17 x dx=d t

−18 −dt
x dx=
17

Using above results, we can write

dx dx −dt
∫ ( 17 ) =∫ 18 ( −17 ) =∫ 17 t
x x +1 x 1+ x

−1
= log t + c
17

= log 1+ 17 + c
( )
−1 1
17 x

Question 9 :-
x
∫ dx
√ x+2+ √ x +1

Solution :-
x
∫ dx
√ x+2+ √ x +1
In such type of question, we have to perform Rationalisation to
remove sq. root in denominator.

dx = ∫ ×√
x+ 2−√ x+ 1
x x
∫ dx
√ x+2+ √ x +1 √ x+2+ √ x +1 √ x+ 2−√ x+ 1

= ∫
x ( √ x+2−√ x +1)
dx
( x+ 2 )−( x+1)

=
1
∫ {[ ( x+ 2 )−2 ] √ x +2−[ ( x+1 ) −1 ] √ x +1 } dx
2−1

Here x = x + 2 – 2 and x = x + 1 – 1
(Extended twice to ease calculation).

=
1
∫¿
1

= ¿

=
2
¿
5
 [ ]
1−tan2 x
Question 10 :-
1
∫ det 2 dx
cos 2 x 2
1+ tan x

[ ]
1−tan 2 ⁡x 1
Solution:- det 2
cos ⁡2 x 2
1+ tan ⁡x

( )
2
1−tan ⁡x
= 2 2
−cos ⁡2 x
1+ tan ⁡x

= 2 cos 2x – cos 2x

= cos 2x

[ ]
1−tan2 x 1
∫ det 2 dx= ∫ cos 2 x dx
cos 2 x 2
1+ tan x

sin ⁡2 x
∴ ∫ cos ⁡2 xdx= +c
2