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Correct ANSWER is (B) \(\frac{e^{2x}}{2} (X^2+x-\frac{35}{4})+C\)

For explanation: By using ∫ U.v dx=u∫ v dx-∫ u'(∫ v dx), we get

\(\int (x^2+9) \,e^x \,dx=(x^2+9) \int e^{2x} \,dx-\int (x^2+9)’\int e^{2x} \,dx\)

=\(\frac{(x^2+9) \,e^{2x}}{2}-\int 2x \frac{e^{2x}}{2} \,dx\)

Again, applying integration by parts for integrating \(\int \,x \,e^{2x} \,dx\)

\(\int \,x \,e^{2x} \,dx=x\int e^{2x} \,dx-\int (x)’ \int \,e^{2x} \,dx\)

=\(\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx\)

=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\).

∴\(\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)

=\(\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})\)

=\(\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C\).

Right OPTION is (c) All value of x

The EXPLANATION: If we solve the L.H.S. (Left Hand Side) of the EQUATION, we GET the FOLLOWING value.

(x+2) (x+4) = x^2 + 4x + 2x + 8 = x^2 + 6x + 8.

This value is same as the R.H.S. (Right Hand Side).

So, all the values of x satisfy the equality.

The correct choice is (d) \(\frac{1}{8} log⁡\left |\frac{x-8}{x+8}\right |+C\)

The best EXPLANATION: \(\INT \frac{2 dx}{x^2-64}=2\int \frac{dx}{x^2-8^2}\)

By USING the formula \(\int \frac{dx}{x^2-a^2}=\frac{1}{2a} log⁡|\frac{x-a}{x+a}|+C\)

∴\(2\int \frac{dx}{x^2-8^2}=2(\frac{1}{(2(8))} log⁡|\frac{x-8}{x+8}|)+2C_1\)

=\(\frac{1}{8} log⁡|\frac{x-8}{x+8}|+C\)

The correct option is (a) Reverse INTEGRAL PROPERTY

To explain I would say: In the reverse integral property the upper limits and lower limits are interchanged. The reverse integral property of definite INTEGRALS is \(\int_a^b\)f(X)DX=-\(\int_b^a\)f(x)dx.

The correct answer is (C) 33

Best explanation: \(I=\int_{-2}^1 \,5x^4 \,dx\)

F(x)=\(\int5x^4 \,dx=5(\FRAC{x^5}{5})=x^5\)

Applying the limits by using the fundamental theorem of calculus, we get

I=F(1)-F(-2)

=(1)^5-(-2)^5=1+32=33.

The correct CHOICE is (b) True

Easiest EXPLANATION: The SUM property of definite integrals is \(\int_a^b\)[F(x)+g(x)dx

\(\int_a^b\)[f(x)+g(x)dx = \(\int_a^b\)f(x)dx+\(\int_a^b\)g(x)dx

Hence, it is true.

Right answer is (C) Identity

For explanation: As it represents the identity (b+a)^2 it SATISFIES the identity (b+a)^2 = (a^2 + b^2 +2ab) and is not linear, cubic or an IMAGINARY equation so the CORRECT option is Identity Equation.

Correct answer is (b) \(\FRAC{8}{3}\) log⁡2-5

To EXPLAIN I would say: \(I=\int_0^1 log⁡x.x^2 dx\)

F(x)=\(\int log⁡x.x^2 dx\)

By using the formula \(\int U.v dx=u\int v dx-\int u'(\int v dx)\), we get

\(\int log⁡x.x^2 \,dx=log⁡x \int x^2 dx-\int (log⁡x)’ \int \,x^2 dx\)

=\(\frac{x^3 log⁡x}{3}-\int \frac{1}{x}.x^3/3 dx\)

∴\(F(x)=\frac{x^3 log⁡x}{3}-\frac{x^3}{9}=\frac{x^3}{3} (log⁡x-\frac{1}{3})\)

Hence, by using the fundamental theorem of calculus, we get

I=F(2)-F(1)

I=\(\frac{2^3}{3} \,(log⁡2-\frac{2}{3})-\frac{1^3}{3} \,(log⁡1-\frac{1}{3})\)

I=\(\frac{2^3}{3} \,log⁡2-\frac{16}{3}+\frac{1}{3}\)

I=\(\frac{8}{3}\) log⁡2-5

Correct CHOICE is (d) \(\frac{11}{9} e^3-\frac{8}{9}\)

The explanation: Let \(I=\int_0^1(x+3) \,e^{3x} \,dx\)

F(x)=\(\int (x+3) \,e^{3x} \,dx\)

By USING the formula \(\int \,u.V \,dx=u\int \,v dx-\int \,u'(\int \,v \,dx)\), we get

F(x)=(x+3) \(\int e^{3x} \,dx-\int \,(x+3)’\int \,e^{3x} \,dx\)

=\(\frac{(x+3) \,e^{3x}}{3}-\int \frac{e^{3x}}{3} dx\)

=\(\frac{(x+3) e^{3x}}{3}-\frac{e^{3x}}{9}\)

=\(\frac{e^{3x}}{3} (x+3-\frac{1}{3})=\frac{e^{3x}}{9}(3x+8)\)

Applying the limits, we get

I=F(1)-F(0)

=\(\frac{e^{3(1)}}{9} (3+8)-\frac{e^{3(0)}}{9}(0+8)\)

=\(\frac{11}{9} e^3-\frac{8}{9}\).

The correct answer is (b) \(\FRAC{1}{6} log \frac{x-3}{x+3}\) + C

The explanation is: \(\INT \frac{dx}{(x^2-9)}=\frac{A}{(x-3)} + \frac{B}{(x+3)}\)

By simplifying, it we get \(\frac{A(x+3)+B(x-3)}{(x^2-9)} = \frac{(A+B)x+3A-3B}{(x^2-9)}\)

By solving the EQUATIONS, we get, A+B=0 and 3A-3B=1

By solving these 2 equations, we get values of A=1/6 and B=-1/6.

Now by putting values in the EQUATION and integrating it we get value,

\(\frac{1}{6} log (\frac{x-3}{x+3})\) + C.

Right option is (a) 2x^4+x+C

Explanation: \(\int \,8X^{3+1} \,dx\)

USING \(\int \,x^n \,dx=\frac{x^{n+1}}{n+1}\), we GET

\(\int \,8x^{3+1} \,dx=\int 8x^3 \,dx+\int \,1 \,dx\)

=\(\frac{8x^{3+1}}{3+1}+x\)

=\(\frac{8x^4}{4}+x\)

=2x^4+x+C.

The CORRECT ANSWER is (d) \(log⁡|x^2+\SQRT{x^2+36}|+C\)

Best EXPLANATION: \(\int \frac{dx}{\sqrt{x^2+36}}\)

By using the formula \(\int \frac{dx}{\sqrt{x^2+a^2}}=log⁡|x^2+\sqrt{x^2+a^2}|+C\)

∴\(\int \frac{dx}{\sqrt{x^2+36}}=log⁡|x^2+\sqrt{x^2+36}|+C\)

Correct OPTION is (a) \(log⁡\left |\frac{4+x}{4-x}\right |+C\)

To elaborate: \(\INT \frac{8dx}{16-x^2}=8\int \frac{dx}{4^2-x^2}\)

By using the formula \(\int \frac{dx}{a^2-x^2}=\frac{1}{2A} \left |\frac{a+x}{a-x}\right |+C\)

∴\(8\int \frac{dx}{4^2-x^2}=8(\frac{1}{2(4)} log⁡\left |\frac{4+x}{4-x}\right |)+8C_1\)

\(8\int \frac{dx}{4^2-x^2}=log⁡\left |\frac{4+x}{4-x}\right |+C\)

The correct OPTION is (c) 2

Best EXPLANATION: \(\int_8^2\)2F(X)dx = -2 \(\int_2^8\)F(x)dx

= – 2(-3)

= 6

The correct ANSWER is (c) 2

Easy EXPLANATION: \(\int_3^2\)f(X)dx = – \(\int_2^3\)f(x)dx

= – 4

Right choice is (a) \(log|X| – \frac{1}{2} log(x^2+1)\) + C

Explanation: We know that \(\int \frac{dx}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}\)

By simplifying it we GET, \(\int \frac{dx}{x(x^2+1)}=\frac{(A+B) x^2+Cx+A}{x(x^2+1)}\)

Now equating the coefficients we get A = 0, B = 0, C=1.

\(\int \frac{dx}{x(x^2+1)} = \int \frac{dx}{x} + \int \frac{-xdx}{(x^2+1)}\)

Therefore after integrating we get \(log|x| – \frac{1}{2} log(x^2+1)\) + C.

The correct answer is (b) 5 log⁡5-4 log⁡4-1

For explanation I WOULD say: LET I=\(\int_4^5 \,log⁡x \,DX\).

F(x)=∫ log⁡x dx

By using the formula \(\int \,u.V dx=u \int v \,dx-\int u'(\int \,v \,dx)\), we get

\(\int log ⁡x \,dx=log⁡x \int \,dx-\int(log⁡x)’\int \,dx\)

F(x)=x log⁡x-∫ dx=x(log⁡x-1).

Applying the LIMITS using the fundamental theorem of calculus, we get

I=F(5)-F(4)=(5 log⁡5-5)-(4 log⁡4-4)

=5 log⁡5-4 log⁡4-1.

The CORRECT choice is (b) \(\frac{9}{2}\)

To explain: I=\(\int_1^4 \frac{\SQRT{x}+3}{\sqrt{x}} \,dx\)

Let \(\sqrt{x}+3=t\)

Differentiating w.r.t x, we get

\(\frac{1}{2\sqrt{x}} \,dx=dt\)

\(\frac{1}{\sqrt{x}} \,dx=2 \,dt\)

The new limits

When x=1,t=4

When x=4,t=5

∴\(\int_1^4 \frac{\sqrt{x}+3}{\sqrt{x}} dx=\int_4^5 \,t \,dt\)

=\([\frac{t^2}{2}]_4^5=\frac{5^2-4^2}{2}=\frac{9}{2}\)

The CORRECT option is (c) \(\frac{X}{2}-\frac{sin⁡(16x+10)}{32}+C\)

The explanation: \(\INT sin^2⁡(8x+5) DX=\int \frac{1-cos⁡2(8x+5)}{2} dx=\int \frac{1}{2} dx-\frac{1}{2} \int cos(16x+10)dx\)

=\(\frac{x}{2}-\frac{1}{2} (\frac{sin⁡(16x+10)}{16})=\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C\)

The correct choice is (B) \(\FRAC{3x}{2}-\frac{sin⁡2x}{4}+C\)

Best explanation: \(\INT \,2 \,sin^2⁡x +cos^2⁡x=\int sin^2⁡x+sin^2⁡x+cos^2⁡x DX\)

=\(\int sin^2⁡x+1 dx\)

=\(\int sin^2⁡x dx+\int 1 dx\)

=\(\int \frac{1-cos⁡2x}{2} dx+\int 1 dx\)

=\(\frac{x}{2}-\frac{sin⁡2x}{4}+x\)

=\(\frac{3x}{2}-\frac{sin⁡2x}{4}+C\)

Right choice is (a) \(\FRAC{3x^4}{2}+18x^2+C\)

To EXPLAIN I would SAY: LET x^2=t

Differentiating w.r.t x, we get

2x dx=dt

\(\int 6x(x^2+6)dx=3\int (t+6) dt\)

3\(\int (t+6)dt=3\left (\frac{t^2}{2}+6t\right )=\frac{3t^2}{2}+18t\)

Replacing t with x^2

\(\int 6x(x^2+6)dx=\frac{3x^4}{2}+18x^2+C\)

The correct ANSWER is (b) \(\FRAC{7 \,sin⁡mx}{m}+C\)

Easy explanation: Using Integration by Substitution, LET xm=t

Differentiating w.r.t x, we get

mdx=dt

∴\(\int 7 \,cos⁡mx \,dx=\int \frac{(7 cos⁡t)}{m} dt\)

=\(\frac{7}{m} \int cos⁡t \,dt=\frac{7}{m} (sin⁡t)+C\)

REPLACING t with mx again we get,

\(\int 7 \,cos⁡mx \,dx=\frac{7 \,sin⁡mx}{m}+C\)

The correct ANSWER is (b) \(3e^x+2 \,log⁡x+\frac{x^4}{4}+C\)

The EXPLANATION: To find \(\INT \,3e^x+\frac{2}{x}+x^3 \,dx\)

\(\int \,3e^x+\frac{2}{x}+x^3 dx=3\int \,e^x \,dx+2\int \frac{1}{x} \,dx+\int x^3 \,dx\)

\(\int \,e^x \,dx=e^x\)

\(\int \frac{1}{x} dx=log⁡x\)

∴\(\int 3e^x+\frac{2}{x}+x^3 \,dx=3e^x+2 \,log⁡x+\frac{x^4}{4}+c\)

Right option is (b) Adding intervals property

For EXPLANATION I WOULD say: The zero-length interval property is one of the PROPERTIES used in definite INTEGRALS and they are always positive. The zero-length interval property is \(\int_a^b\)f(X)dx = 0.

Right option is (b) \(\frac{124}{3}\)

To elaborate: Let I=\(\int_1^5x^2 \,DX\)

F(x)=\(\INT x^2 \,dx\)

=\(\frac{x^3}{3}\)

By using the fundamental theorem of calculus, we get

I=F(5)-F(1)

=\(\frac{5^3}{3}-\frac{1^3}{3}=\frac{125-1}{3}=\frac{124}{3}\)

The correct ANSWER is (C) e^3-1

Explanation: Let I=\(\int_0^3 \,e^x \,DX\)

F(x)=\(\int \,e^x \,dx=e^x\)

Applying the LIMITS, we get

I=F(3)-F(0)

=e^3-e^0=e^3-1.

The correct ANSWER is (b) 9(e^6 – e^3)

The best EXPLANATION: \(\int_3^6\)9 e^x dx = 9(e^x)^63 dx

= 9(e^6 – e^3)

Correct option is (b) \(tan^{-1}⁡\FRAC{x}{3}+C\)

The best I can explain: \(\int \frac{3dx}{9+x^2}=3\int \frac{DX}{3^2+x^2}\)

Using the FORMULA \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}\frac{⁡x}{a}+C\)

∴\(3\int \frac{dx}{x^2+3^2}=3\left (\frac{1}{3} tan^{-1}⁡\frac{x}{3}\right )+3C_1\)

\(3\int \frac{dx}{x^2+3^2}=tan^{-1}⁡\frac{x}{3}+C\).

Right choice is (d) -log⁡(1-sin⁡2x)+C

The explanation is: \(\int \frac{2 cos⁡2x}{(cos⁡X-sin⁡x)^2} DX=\int \frac{2 cos⁡2x}{cos^2⁡x+sin^2⁡x-sin⁡2x} \,dx \,(∵2 cos⁡x sin⁡x=sin⁡2x)\)

=\(\int \frac{2 cos⁡2x}{1-sin⁡2x} dx\)

Let 1-sin⁡2x=t

Differentiating w.r.t x, we get

-2 cos⁡2x dx=DT

2 cos⁡2x dx=-dt

\(\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-\int \frac{dt}{t}\)

=-log⁡t

Replacing t with 1-sin⁡2x, we get

∴\(\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}\) dx=-log⁡(1-sin⁡2x)+C

Right CHOICE is (C) –\(\FRAC{1}{7}\) (cot⁡x+tan⁡x)+C

Explanation: To find: \(\int \frac{cos^2⁡x-sin^2⁡x}{7 \,cos^2⁡x \,sin^2⁡x} DX\)

\(\int \frac{cos^2⁡x-sin^2⁡x}{7 \,cos^2⁡x \,sin^2⁡x} dx=\frac{1}{7} \int \frac{1}{sin^2⁡x}-\frac{1}{cos^2⁡x} dx\)

=\(\frac{1}{7} \int cosec^2 x-sec^2⁡x dx\)

=\(\frac{1}{7}\) (-cot⁡x-tan⁡x)+C

=-\(\frac{1}{7}\) (cot⁡x+tan⁡x)+C.

Right answer is (c) \(\frac{9}{4}\LEFT (\frac{π}{2}-1\right)\)

For explanation: LET I=\(\int_0^{\frac{π}{4}} \,9 \,cos^2⁡x \,dx\).

F(x)=\(\int \,9 \,cos^2⁡x \,dx\)

=9\(\int(\frac{1+cos⁡2x}{2})dx\)

=\(\frac{9}{2} (x-\frac{sin⁡2x}{2})\)

APPLYING the limits, we get

I=\(F(\frac{π}{4})-F(0)=\frac{9}{2} \left (\frac{π}{4}-\frac{sin⁡2(\frac{π}{4})}{2}\right)-\frac{9}{2} (0-\frac{sin⁡0}{2})\)

=\(\frac{9}{2}\left (\frac{π}{4}-\frac{sin⁡π/2}{2}\right )=\frac{9}{4} (π/2-1)\)

The CORRECT choice is (C) 6

To elaborate: \(\int_3^7\)2F(x)-g(x)dx = 2 \(\int_3^7\)f(x)– \(\int_3^7\)g(x)dx

= 2(4) – 2

= 6

Right ANSWER is (a) 7(1-\(\frac{1}{\sqrt{2}}\))

To explain I would say: LET \(I=\int_{π/4}^{π/2}7 \,cos⁡x \,dx\)

F(x)=∫ 7 cos⁡x dx

=7(sin⁡x)

Applying the limits by USING the second fundamental theorem of calculus, we GET

\(I=F(\frac{π}{2})-F(\frac{π}{4})=7(sin\frac{π}{2}-sin⁡ \frac{π}{4})=7(1-\frac{1}{\sqrt{2}})\)

Correct answer is (B) ADDING intervals PROPERTY

The explanation is: \(\int_a^b\)f(x)dx+\(\int_b^c\)(x)dx = \(\int_a^c\)f(x) dx is a property of definite integrals. \(\int_a^b\)f(x)dx+\(\int_b^c\)f(x)dx = \(\int_a^c\)f(x) dx is called as adding intervals property USED to COMBINE a lower limit and upper limit of two different integrals.

Right answer is (c) \(\int_a^b\)k⋅f(X)dx

Easiest explanation: The CONSTANT multiple property of definite INTEGRALS is \(\int_a^b\)k⋅f(x)dx

\(\int_a^b\)k⋅f(x)dx = k \(\int_a^b\)f(x)dx

The correct answer is (C) 10 \( log⁡|x+\SQRT{x^2-25}|+C\)

Best explanation: \(\INT \FRAC{10 \,dx}{\sqrt{x^2-25}}=10\int \frac{dx}{\sqrt{x^2-25}}\)

By using the formula \(\int \frac{dx}{\sqrt{x^2-a^2}}=log⁡|x+\sqrt{x^2-a^2}|+C\), we get

∴\(10 \int \frac{dx}{\sqrt{x^2-25}}=10 log⁡|x+\sqrt{x^2-25}|+10C_1\)

=10 \(log⁡|x+\sqrt{x^2-25}|+C\)

The CORRECT CHOICE is (B) 2.1

Easiest EXPLANATION: \(\int_1^2\)1y^5/5dy = \(\frac {1}{5}\)(y^6/6)^21

= \(\frac {1}{5}(\frac {64}{6} – \frac{1}{6})\)

= 2.1

Right answer is (a) 14(3 log⁡2-1)

EXPLANATION: I=\(\int_{\sqrt{2}}^2 \,14X \,log⁡ x^2 \,DX\)

Let x^2=t

Differentiating w.r.t x, we get

2x dx=dt

The new limits

When x=\(\sqrt{2}\), t=2

When x=2, t=4

∴\(\int_{\sqrt{2}}^2 \,14x \,log⁡ x^2 \,dx =\int_2^4 \,7 \,log⁡ t \,dt\)

Using INTEGRATION by parts, we get

\(\int_2^4 \,7 \,log⁡ t \,dt=7(log⁡ t\int dt-\int (log⁡t)’ \int \,dt)\)

=7 (t log⁡t-t)2^4

=7(4 log⁡4-4-2 log⁡2+2)

=7(6 log⁡2-2)=14(3 log⁡2-1)

The correct CHOICE is (b) True

Explanation: In \(\int_a^b\)f(y) dy ‘a’ is the called as lower limit and ‘b’ is called the UPPER limit of the integral. The fuction ‘f’ in \(\int_a^b\)f(y)dy is called the INTEGRAND. The letter ‘y’ is a dummy symbol and can be REPLACED by any other symbol.

Right option is (a) \(SIN^{-1}⁡\FRAC{x}{\SQRT{5}}+C\)

Easy explanation: \(\int \frac{dx}{\sqrt{5-x^2}}=\int \frac{dx}{\sqrt{(√5)^2-x^2}}\)

By USING the formula \(\int \frac{dx}{\sqrt{a^2-x^2}}=sin^{-1}⁡\frac{x}{a}+C\)

∴\(\int \frac{dx}{\sqrt{x^2-5}}=sin^{-1}⁡\frac{x}{\sqrt{5}}+C\)

The correct ANSWER is (a) \(Log \left|\frac{x+1}{x+2}\right|+ C\)

The best I can explain: It is a proper rational FUNCTION. Therefore,

\(\frac{1}{(x+1)(x+2)} = \frac{A}{(x+1)} + \frac{B}{(x+2)}\)

Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1. SOLVING it we get A=1, and B=-1.

Thus, it simplifies to, \(\frac{1}{(x+1)} + \frac{-1}{(x+2)} = \int \frac{DX}{(x+1)} – \int \frac{dx}{(x+2)}\).

= log|x+1| – log|x+2| + C

= \(Log \left|\frac{x+1}{x+2}\right|+ C\).