1. (b)
$e^{5ln x} = e^ {ln x^5} =x^5$
So
$\int e^{5ln x}\; dx= \int x^5 \; dx= \frac {x^6}{6} + C $
2. (a) \( \ln |x| + C \)
3. (a) \( e^x + C \)
4. (c) \( \frac{3}{2}x^3 + C \)
Use the power rule for integration: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\)
5. (a) \( \tan x + C \)
6. (d) -2
$\int_{-1}^{1} \frac {|x-2|}{x-2} \; dx = \int_{-1}^{1} \frac {-(x-2)}{x-2} \; dx= \int_{-1}^{1} -1 \; dx=-2$
7. (a) \( \sin^{-1} x + C \)
8. (a) \( \tan^{-1} x + C \)
9. (a) \( 2\sqrt{x} + C \)
Use the power rule. \( \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}} \), so the integral is \( \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} = 2\sqrt{x} \).
10. (c) \( \frac{x}{2} - \frac{\sin 2x}{4} + C \)
Use the half-angle identity: \( \sin^2 x = \frac{1}{2}(1 - \cos 2x) \). Integrating this gives \( \frac{1}{2}x - \frac{1}{4}\sin 2x \).
11. (a) \( \frac{2^x}{\ln 2} + C \)
The integral of \( a^x \) is \( \frac{a^x}{\ln a} \), so for \( 2^x \), it is \( \frac{2^x}{\ln 2} \).
12. (a) \( x \ln x - x + C \)
By integration by parts, \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x \).
13. (a) \( -\ln |\cos x| + C \)
14. (a) \( \tan x + C \)
15. (a)
Integration by substitution
$t=x^3$
$dt=3x^2 dx$
$x^2 dx= \frac {1}{3} dt$
$\int x^2 e^{x^3} \; dx= \frac {1}{3} \int e^{t} \; dt= \frac {1}{3} e^t + C= \frac {1}{3} e^{x^3} + C$
16. (c) \( \frac{x}{2} + \frac{\sin 2x}{4} + C \)
Use the half-angle identity: \( \cos^2 x = \frac{1}{2}(1 + \cos 2x) \).
Integrating this gives \( \frac{1}{2}x + \frac{1}{4}\sin 2x \).
17. (d)
Integration by substitution
$t=x^2 + 1$
$dt=2x dx$
$x dx= \frac {1}{2} dt$
$\int_{2}^{4} \frac {x}{x^2 + 1} \; dx= \frac {1}{2} \int_{5}^{17} \frac {1}{t} \; dt= [\frac {1}{2} ln |t| ]_{5}^{17}= \frac {1}{2} log \frac {17}{5}$
18. (c) 1/2
$\int_{0}^{\pi/4} \sin 2x \; dx= [-\frac {1}{2} cos 2x]_{0}^{\pi/4}=\frac {1}{2}$
19. (a) \( \ln |x| - \ln |x+1| + C \)
Use partial fractions: \( \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \). Solving for A and B gives \( A = 1 \) and \( B = -1 \). Integrating gives \( \ln |x| - \ln |x+1| \).
20. (a) \( -\sqrt{1 - x^2} + C \)
Use substitution: Let \( u = 1 - x^2 \). Then \( du = -2x dx \). The integral becomes \( -\frac{1}{2} \int \frac{du}{\sqrt{u}} \), which is \( -\sqrt{u} \) or \( -\sqrt{1 - x^2} \).
Apply integration by parts:
\[ \int \sec^3 x \, dx = \int \sec x \cdot \sec^2 x \, dx = \sec x \tan x - \int \tan x \cdot \sec x \tan x \, dx \]
\[ = \sec x \tan x - \int \sec x \tan^2 x \, dx \]
Now, use the trigonometric identity \( \tan^2 x = \sec^2 x - 1 \):
\[ = \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx \]
\[ = \sec x \tan x - \int \sec^3 x \, dx + \int \sec x \, dx \]
Notice that \( \int \sec^3 x \, dx \) appears on both sides of the equation. Let's denote this integral as \( I \):
\[ I = \sec x \tan x - I + \int \sec x \, dx \]
Adding \( I \) to both sides gives:
\[ 2I = \sec x \tan x + \int \sec x \, dx \]
Now, \( \int \sec x \, dx \) is a standard integral, equal to \( \ln | \sec x + \tan x | + C \). So,
\[ 2I = \sec x \tan x + \ln | \sec x + \tan x | + C \]
Therefore
\[ I = \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln | \sec x + \tan x | + C' \]
To solve the integral \( \int e^{2x} \sin(3x+1) \, dx \), we can use integration by parts twice. Integration by parts is based on the formula:
\[ \int u \, dv = uv - \int v \, du \]
For this integral, we'll apply the method twice. Let's start with the first application:
(a).First Application:
- Choose \( u = \sin(3x+1) \) and \( dv = e^{2x} \, dx \).
- Then, \( du = 3\cos(3x+1) \, dx \) and \( v = \frac{1}{2}e^{2x} \).
Applying integration by parts:
\[ \int e^{2x} \sin(3x+1) \, dx = \frac{1}{2}e^{2x} \sin(3x+1) - \frac{3}{2} \int e^{2x} \cos(3x+1) \, dx \]
(b). Second Application:
- For the remaining integral, choose \( u = \cos(3x+1) \) and \( dv = e^{2x} \, dx \).
- Then, \( du = -3\sin(3x+1) \, dx \) and \( v = \frac{1}{2}e^{2x} \).
Applying integration by parts again:
\[ \int e^{2x} \cos(3x+1) \, dx = \frac{1}{2}e^{2x} \cos(3x+1) + \frac{3}{2} \int e^{2x} \sin(3x+1) \, dx \]
Now, substitute this back into our first equation:
\[ \int e^{2x} \sin(3x+1) \, dx = \frac{1}{2}e^{2x} \sin(3x+1) - \frac{3}{4}e^{2x} \cos(3x+1) - \frac{9}{4} \int e^{2x} \sin(3x+1) \, dx \]
Let \( I = \int e^{2x} \sin(3x+1) \, dx \). Then, we have:
\[ I = \frac{1}{2}e^{2x} \sin(3x+1) - \frac{3}{4}e^{2x} \cos(3x+1) - \frac{9}{4}I \]
\[ \frac{13}{4}I = \frac{1}{2}e^{2x} \sin(3x+1) - \frac{3}{4}e^{2x} \cos(3x+1) \]
\[ I = \frac{2}{13}e^{2x} \sin(3x+1) - \frac{3}{13}e^{2x} \cos(3x+1) + C \]
$=\int \frac {\sin(x+a -2a)}{\sin(x+a)} \; dx$
$=\int \frac {\sin(x+a)\cos2a -\cos(x+a) \sin 2a}{\sin(x+a)} \; dx$
$=\int \cos2a - \frac {\cos(x+a) \sin 2a}{\sin(x+a)} \; dx$
put $\sin(x+a)=t$
$\cos (x+a) dx dt$
$=\int \cos2a \; dx - \sin 2a \int \frac {1}{t} \; dt$
$=x \cos2a - \sin 2a ln |\sin(x+a)| + C$
To integrate \( \int_{0}^{3/2} |x \sin \pi x| \, dx \), we need to consider the behavior of the function \( x \sin \pi x \) over the interval \([0, \frac{3}{2}]\) and where it changes sign, since the absolute value affects the integral.
Within the interval \([0, \frac{3}{2}]\), \( \sin \pi x \) is zero at \( x = 0 \) and \( x = 1 \), and changes sign at these points. Therefore, we need to split the integral at \( x = 1 \):
1. From \( x = 0 \) to \( x = 1 \), \( \sin \pi x \) is positive, so \( |x \sin \pi x| = x \sin \pi x \).
2. From \( x = 1 \) to \( x = \frac{3}{2} \), \( \sin \pi x \) is negative, so \( |x \sin \pi x| = -x \sin \pi x \).
Therefore we can write the integral as:
\[ \int_{0}^{3/2} |x \sin \pi x| \, dx = \int_{0}^{1} x \sin \pi x \, dx + \int_{1}^{3/2} -x \sin \pi x \, dx \]
Let's solve these integrals separately:
1.First Integral \( \int_{0}^{1} x \sin \pi x \, dx \):
Applying integration by parts:
\[ \int x \sin \pi x \, dx = -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi} \int \cos \pi x \, dx \]
\[ = -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x + C \]
Evaluate this from 0 to 1:
\[ \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_{0}^{1} \]
\[ = \frac{1}{\pi} \]
2. Second Integral \( \int_{1}^{3/2} -x \sin \pi x \, dx \):
Again Applying integration by parts:
\[ \left[ -\frac{x}{\pi} \cos \pi x + \frac{1}{\pi^2} \sin \pi x \right]_{1}^{3/2} \]
\[ = \left( -\frac{3}{2\pi} \cos \frac{3\pi}{2} + \frac{1}{\pi^2} \sin \frac{3\pi}{2} \right) - \left( -\frac{1}{\pi} \cos \pi + \frac{1}{\pi^2} \sin \pi \right) \]
\[ = \left( -\frac{3}{2\pi} \cdot 0 - \frac{1}{\pi^2} \cdot (-1) \right) - \left( -\frac{1}{\pi} \right) \]
\[ = \frac{1}{\pi^2} + \frac{1}{\pi} \]
Adding these two results together gives the total integral:
\[ \int_{0}^{3/2} |x \sin \pi x| \, dx = +\frac{1}{\pi} + \left( \frac{1}{\pi^2} + \frac{1}{\pi} \right) \]
\[ = \frac {2}{\pi} + \frac{1}{\pi^2} \]
Answer is 23/3
Let
$\frac {x^2 + x +1}{(x+1)^2 (x+2)} = \frac {A}{x+1} + \frac {B}{(x+1)^2} + \frac {C}{x+2}$
or
$x^2 + x +1= A(x+1)(x+2) + B(x+2) + C(x+1)^2$
At x=-1, we get
$B=1$
At x=-2, we get
C=3
At x=0, we get
A=-2
Therefore
$\frac {x^2 + x +1}{(x+1)^2 (x+2)} = \frac {-2}{x+1} + \frac {1}{(x+1)^2} + \frac {3}{x+2}$
So,
$\int \frac {x^2 + x +1}{(x+1)^2 (x+2)} \; dx= \int \frac {-2}{x+1} + \frac {1}{(x+1)^2} + \frac {3}{x+2} \; dx $
$= -2 log |x+1| - \frac {1}{x+1} + 3log |x+2| + C$
Let
$\frac {x}{(x^2+1) (x-2)} = \frac {Ax+B}{x^2 + 1} + \frac {C}{x-1}$
or
$x= (Ax+B)(x-1) + C(x^2 +1)$
Comparing coefficent and solving ,we get
A=-1/2, B=1/2 and C=1/2
Therefore
$\frac {x}{(x^2+1) (x-2)} = \frac {-1/2(x-1)}{x^2 + 1} + \frac {1/2}{x-1}$
So,
$\int \frac {x}{(x^2+1) (x-2)} \; dx$
$=\int \frac {-1/2(x-1)}{x^2 + 1} + \frac {1/2}{x-1} \; dx $
$=\frac {1}{2} \int \frac {-x}{x^2 + 1} + \frac {1}{x^2 + 1} + \frac {1}{x-1} \; dx$
$= \frac {1}{2}[ \frac {-1}{2} log |x^2 + 1| + \tan^{-1} x + log |x-1| ] +C$
(a)
Let
$\frac{2x + 3}{(x - 1)(x + 2)} = \frac {A}{x-1} + \frac {B}{(x+2)}$
$2x+3 = A(x+2) + B(x-1)$
or A= 5/3
B=1/3
So,
$\frac{2x + 3}{(x - 1)(x + 2)} =\frac {1}{3}[ \frac {5}{x-1} + \frac {1}{(x+2)}]$
Therefore
$\int \frac{2x + 3}{(x - 1)(x + 2)} \; dx= \frac {1}{3} \int [ \frac {5}{x-1} + \frac {1}{(x+2)}] \; dx$
$=\frac {1}{3} [ 5 \ln |x-1| + \ln |x+2| ] + C$
Let $I =\int _{0}^{\pi/2} \frac {x\sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)} \; dx$ -(1)
Using
$\int _{0}^{a} f(x) dx = \int _{0}^{a} f(a-x) dx $, we get
$I= \int _{0}^{\pi/2} \frac {(\frac {\pi}{2} -x) \sin (\frac {\pi}{2} -x) \cos(\frac {\pi}{2} -x)}{\sin^4 (\frac {\pi}{2} -x) + \cos^4 (\frac {\pi}{2} -x)} \; dx$
$I= \int _{0}^{\pi/2} \frac {(\frac {\pi}{2} -x) \sin (x) \cos(x)}{\sin^4 (x) + \cos^4 x)} \; dx$ -(2)
Adding (1) and (2)
$2I=\frac {\pi}{2} \int _{0}^{\pi/2} \frac {\sin(x) \cos(x)}{\sin^4(x) + \cos^4(x)} \; dx$
Dividing numerator and denominator by $\cos^4 x$, we get
$2I=\frac {\pi}{2} \int _{0}^{\pi/2} \frac {\tan (x) \sec^2(x)}{1+ \tan^4 (x)} \; dx$
Put $\tan^2 x= t$, then $2\tan (x) \sec^2(x) dx =dt$
Therefore
$I = \frac {\pi}{8} \int _{0}^{\infty} \frac {1}{1+t^2} \; dt $
$I= \frac {\pi}{8} [ tan^{-1} t] _{0}^{\infty}= \frac {\pi^2}{16}$