\( \mathrm{I}=\int x^{3} e^{x} d x \)
\( \mathrm{u}=x^{3} \quad \int d v=\int e^{x} d x \)
\(d u=3 x^{2} d x \quad v=e^{x} \)
\( \int u d v=u v-\int u d v \text { (or) } \)
\( \mathrm{I}=\mathrm{uv}-u^{\prime} v_{1}+u^{\prime} v_{2}-u^{\prime \prime} v_{3} \)
\( \mathrm{u}=x^{3}, \quad \int d v=\int e^{x} d x \)
\(u^{\prime}=3 x^{2}, \quad v=e^{x}, \)\(u^{\prime \prime}=6 x \quad \mathrm{v}_{1}=\mathrm{e}^{\mathrm{x}} \)
\( u^{\prime \prime}=6, \quad \mathrm{v}_{2}=\mathrm{e}^{\mathrm{x}} \ \)
\(v_{3}=e^{x} \)
\( I=x^{3} e^{x}-3 \int e^{x} x^{2} d x \)
\( =x^{3} e^{x}-3 e^{x} x^{2}+6 x e^{x}-6 e^{x}+c\)

\(\mathrm{I}=\int \frac{x^{3}+x}{x^{4}-9} d x=\int \frac{x^{3}}{x^{4}-9} d x+\int \frac{x d x}{x^{4}-9}\)
\(=I_{1}+I_{2}\) ............ (1)
\( \mathrm{I}_{1} =\int \frac{x^{3}}{x^{4}-9} d x \quad t=x^{4}-9, \mathrm{dt}=4 x^{3} \mathrm{~d} x \)
\( =\frac{1}{4} \int \frac{d t}{t}=\frac{1}{4} \log |t| \frac{d t}{4}=x^{3} \mathrm{~d} x \)
\( =\frac{1}{4} \log \left|x^{4}-9\right|\)
\( \mathrm{I}_{2} =\int \frac{x d x}{x^{4}-9} \)
\( =\int \frac{x}{\left(x^{2}\right)^{2}-3^{2}} d x \)
\(\mathrm{u} =x^{2}, \mathrm{du}=2 x \mathrm{~d} x \)
\( =\frac{1}{2} \int \frac{d u}{u^{2}-3^{2}} \)
\(=\frac{1}{2} \cdot \frac{1}{2(3)} \log \left|\frac{u-3}{u+3}\right| \)
\( =\frac{1}{12} \log \left|\frac{x^{2}-3}{x^{2}+3}\right|\)
Substitute in (1)
\(I=\frac{1}{4} \log \left|x^{4}-9\right|+\frac{1}{12} \log \left|\frac{x^{2}-3}{x^{2}+3}\right|+c\)

\( \int \frac{1}{3 x^{2}+13 x-10} d x \)
\( =\frac{1}{3} \int \frac{d x}{x^{2}+\frac{13 x}{3}-\frac{10}{3}} \)
\(=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\frac{10}{3}-\frac{169}{36}} \)
\(=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\frac{289}{36}} \)
\( =\frac{1}{17} \log \left|\frac{1}{x+5}\right|+c \)
\( =\frac{1}{3} \int \frac{4 x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}} \)
\( =\frac{1}{3} \cdot \frac{1}{2\left(\frac{17}{6}\right)} \log \left|\frac{13}{x+\frac{13}{6}+\frac{17}{6}}\right|+c\)
\( =\frac{1}{17} \log \left|\frac{x-\frac{4}{6}}{x+5}\right|+c \)
\( =\frac{1}{17} \log \left|\frac{3 x-2}{3(x+5)}\right|+c \)

\(\int \frac{x^{3}}{(x+1)^{2}} d x=\int \frac{x^{3}}{x^{2}+2 x+1} d x\)

\( \frac{x^{3}}{(\mathrm{x}+1)^{2}} =x-2+\frac{3 x+2}{(\mathrm{x}+1)^{2}} \)
\(\int \frac{x^{3}}{(\mathrm{x}+1)^{2}} d x =\int(\mathrm{x}-2) \mathrm{d} \mathrm{x}+\int \frac{(3 \mathrm{x}+2)}{(\mathrm{x}+1)^{2}} d x \)
\(=\int(x-2) d x+I\) ............... (i)
Where \(I=\int \frac{(3 x+2)^{0}}{(x+1)^{2}} d x\)
\( \frac{3 \mathrm{x}+2}{(\mathrm{x}+1)^{2}} =\frac{A}{\mathrm{x}+1}+\frac{B}{(\mathrm{x}+1)^{2}} \)
\(3 x+2 =\mathrm{A}(x+1)+\mathrm{B} \)
\(\text {If } x =-1 \)
\(-1 =\mathrm{B} \)
\(\text {If } x =0 \)
\(2 =\mathrm{A}+\mathrm{B} \Rightarrow \mathrm{A}=3 \)
\(\text {I } =\int \frac{3}{x+1} d x-\int \frac{d x}{(x+1)^{2}} \)
\(=3 \log (x+1)-\int(x+1)^{-2} d x \)
\(=3 \log (x+1)-\frac{(x+1)^{-1}}{-1}+c \)
\(=3 \log (x+1)+\frac{1}{x+1}+c\)
Substitute in (1)
\( =\int \frac{x^{3}}{(x+1)^{2}} d x \)
\( =\frac{x^{2}}{2}-2 x+3 \log |x+1|+\frac{1}{x+1}+c\)

\( \int\left[\frac{2+x+x^{2}}{x^{2}(2+\mathrm{x})}+\frac{2 x-1}{(\mathrm{x}+1)^{2}}\right] d x \)
\( =\int\left[\frac{2+x}{x^{2}(2+\mathrm{x})}+\frac{x^{2}}{\mathrm{x}^{2}(2+x)}+\frac{2(\mathrm{x}+1)-3}{(\mathrm{x}+1)^{2}}\right] d x \)
\( =\int\left[\frac{1}{x^{2}}+\frac{1}{2+x}+\frac{2}{\mathrm{x}+1}-\frac{3}{(\mathrm{x}+1)^{2}}\right] d x \)
\(=\int\left[x^{-2}+\frac{1}{2+x}+\frac{2}{x+1}-3(\mathrm{x}+1)^{-2}\right] d x \)
\(=\frac{x^{-1}}{-1}+\log |2+x|+2 \log |x+1|-\frac{3(x+1)^{-1}}{-1}+c \)
\(=\frac{-1}{x}+\log |2+x|+2 \log |\mathrm{x}+1|+\frac{3}{(\mathrm{x}+1)}+c\)

Let I = \(\int _{ 2 }^{ 4 }{ (2x-1) } dx\)
Here a = 2, b = 4 ⇒ \(h=\frac { b-a }{ n } =\frac { 4-2 }{ n } =\frac { 2 }{ n } \)
Given f(x) = 2x - 1
\(f(a+rh)=f\left( 2+r.\frac { 2 }{ n } \right) \)
= \(2\left( 2+\frac { 2r }{ n } \right) -1\)
= \(4+\frac { 4r }{ n } -1=3+\frac { 4r }{ n } \)
\(\therefore f(a+rh)=3+\frac { 4r }{ n } \)
\(\therefore \int _{ 2 }^{ 4 }{ (2x-1) } =\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 2 }{ n } } } \left( 3+\frac { 4r }{ n } \right) \)
= \(\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 2 }{ n } } } \left( 3+\frac { 4r }{ n } \right) \)
= \(\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \left( \frac { 6 }{ n } +\frac { 8r }{ { n }^{ 2 } } \right) } } \)
= \(\lim _{ n\rightarrow \infty }{ \left( \frac { 6 }{ n } .\sum _{ r=1 }^{ n }{ .1 } +\frac { 8 }{ { n }^{ 2 } } .\sum _{ r=1 }^{ n }{ r } \right) } \)
= \(\lim _{ n\rightarrow \infty }{ \left( \frac { 6 }{ n } .n+\frac { 8 }{ { n }^{ 2 } } .\frac { n(n+1) }{ 2 } \right) } \)

= \(6+4\lim _{ n\rightarrow \infty }{ \left( 1+\frac { 1 }{ n } \right) } \)
= \(\therefore \int _{ 2 }^{ 4 }{ (2x-1) } dx=10\)
= \(\left[ \because \sum _{ r=1 }^{ n }{ r } =n\& \sum _{ r=1 }^{ n }{ r } n=\frac { n(n+1) }{ 2 } \right] \)

Let I = \(\int _{ a }^{ b }{ \frac { f\left( x \right) }{ f\left( x \right) +f(a+b-x) } } dx=\frac { b-a }{ 2 } --(1)\)
By the property, \(\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f(a+b-x) } dx\)
\(I=\int _{ a }^{ b }{ \frac { f(a+b-x) }{ f(a+b-x)+f(a+b-(a+b-x)) } } dx\)
= \(\int _{ a }^{ b }{ \frac { f(a+b-x) }{ (a+b-x)+f(a+b-a-b+x) } } dx\)
= \(\int _{ a }^{ b }{ \frac { f(a+b-x) }{ f(a+b-x)+f(x) } } ---(2)\)
Adding (1) and (2) we get, 2I= \(\int _{ a }^{ b }{ \left( \frac { f(x) }{ f(x)+f(a+b-x) } +\frac { f(a+b-x) }{ f(a+b-x)+f(x) } \right) } \)

= \(\int _{ a }^{ b }{ dx } ={ \left[ x \right] }_{ a }^{ b }=b-a\)
2I = b - a
\(\Rightarrow I=\frac { b-a }{ 2 } \) Hence proved

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin } 3x\ sin\ 2x\ dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ 2sin } 3xsin2x\quad dx\)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left[ cos(3x-2x)-cos(3x+2x) \right] } dx\)
\(\left[ \because 2sin(sinD=cos(C-D))-cos(C+D) \right] \)
= \(\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (cosx-cos5x)dx } \)
= \(\frac { 1 }{ 2 } { \left[ sinx-\frac { sin5x }{ 5 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
= \(\frac { 1 }{ 2 } \left[ \left( sin\frac { \pi }{ 4 } -\frac { 1 }{ 5 } sin5\frac { \pi }{ 4 } \right) -\left( sin0-\frac { 1 }{ 5 } sin5(0) \right) \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ 5 } \left( \frac { -1 }{ \sqrt { 2 } } \right) \right] \quad \quad \left[ \because sin0=0 \right] \)

= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 5\sqrt { 2 } } \right] \)
= \(\frac { 1 }{ 2 } \left( \frac { 5+1 }{ 5\sqrt { 2 } } \right) =\frac { 1 }{ 2 } \times \frac { 6 }{ 5\sqrt { 2 } } =\frac { 3 }{ 5\sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \)
= \(\frac { 3\sqrt { 2 } }{ 10 } \)

Let I = \(\int { \frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } } \)
Consider
\(\frac { { x }^{ 2 }+1 }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\frac { A }{ x-1 } +\frac { B }{ { \left( x-1 \right) }^{ 2 } } +\frac { C }{ x+3 } \quad ----(1)\)
\(\frac { { x }^{ 2 }+1 }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\quad \frac { A(x-1)(x+3)+B(x+3)+C{ (x-1) }^{ 2 } }{ { (x-1) }^{ 2 }(x+3) } \)
⇒ x² + 1 = A (x - 1)(x+3) + B (x + 3) + C (x-1)²
When X = 1
1 + 1 = B (1 + 3) ⇒ 2 = b (4)
⇒ \(B=\frac { 2 }{ 4 } \Rightarrow B=\frac { 1 }{ 2 } \)
When x = -3
(-3)² + 1 = C (-4)² ⇒ 10 = 16C
⇒ \(C=\frac { 10 }{ 16 } \)
\(C=\frac { 5 }{ 8 } \)
When x = 0,
1 = -3A +3B + 3C
\(1=-3A+3\left( \frac { 1 }{ 2 } \right) +\frac { 5 }{ 8 } \)
\(3A=-1\frac { 3 }{ 2 } +\frac { 5 }{ 8 } =\frac { -8+12+5 }{ 8 } \)
= \(\frac { 9 }{ 8 } \)
\(A=\frac { 9 }{ 8 } \)
From (1), \(\frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } =\frac { \frac { 3 }{ 8 } }{ x-1 } +\frac { \frac { 1 }{ 2 } }{ { (x-1) }^{ 2 } } +\frac { \frac { 5 }{ 8 } }{ x+3 } \)
\(I=\frac { \left( { x }^{ 2 }+1 \right) dx }{ { \left( x-1 \right) }^{ 2 }\left( x+3 \right) } \)
= \(\frac { 3 }{ 8 } \int { \frac { dx }{ x-1 } + } \frac { 1 }{ 2 } \int { \frac { dx }{ { (x-1) }^{ 2 } } } +\frac { 5 }{ 8 } \int { \frac { dx }{ x+3 } } \)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| +\frac { 1 }{ 2 } \int { { (x-1) }^{ -2 } } dx+\frac { 5 }{ 8 } log\left| x+3 \right| +c\)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| +\frac { 1 }{ 2 } \frac { { (x-1) }^{ -1 } }{ -1 } +\frac { 5 }{ 8 } log\left| x+3 \right| +c\)
= \(\frac { 3 }{ 8 } log\left| x-1 \right| -\frac { 1 }{ 2(x-1) } +\frac { 5 }{ 8 } log\left| x+3 \right| +c\)

Let I = [sin (log x) + cos (log x)] dx
Put log x = t ⇒ x = e^t
⇒ dx = e^t. dt
I = ഽ(sin t + cos t). e^t dt
= ഽ e^t (sin t + cos t) dt
Let f(t) = sin t ⇒ f' (t) = cos t
I = ഽe^t (f (t) + f' (t)) dx
= e^t . f(t) +c
[∵ ഽ e^t [ (f (t) +f' (t) ] dt = e^t . f (t) +c]
= e^{log x}. sin (log x) +c
= x sin (log x) +c   [∵ e^{log x} =x]

Let I = ഽx. log (1 + x) dx

u = log (1 + x); dv = x dx
\(du=\frac { 1 }{ 1+x } dx;\quad v=\frac { { x }^{ 2 } }{ 2 } \)
Using integration by parts we get,
I = ഽu dv = uv - ഽv du
= ഽ x log (1 + x) dx
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\int { \frac { { x }^{ 2 } }{ 2 } .\frac { 1 }{ 1+x } } dx\)
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \int { \frac { { x }^{ 2 } }{ 1+x } } dx\)
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \int { \frac { { x }^{ 2 }-1+1 }{ 1+x } } dx\)
[Adding & subtracting 1 in the numerator]
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \)
 
= \(\frac { { x }^{ 2 } }{ 2 } log(1+x)-\frac { 1 }{ 2 } \left[ \int { \left( x-1 \right) dx+log\left| 1+x \right| } \right] +c\)
= \(\frac { { x }^{ 2 } }{ 2 } log(x+1)-\frac { 1 }{ 2 } \left[ \frac { { x }^{ 2 } }{ 2 } -x+log\left| 1+x \right| \right] +c\)

Let I = \(\int { \frac { 1 }{ { 3x }^{ 2 }+13x-10 } } dx\)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x-\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { x }^{ 2 }+\frac { 13 }{ 3 } x+\frac { 169 }{ 36 } -\frac { 169 }{ 36 } -\frac { 10 }{ 3 } } } \)
= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-\frac { 289 }{ 36 } } } \)

= \(\frac { 1 }{ 3 } \int { \frac { dx }{ { \left( x+\frac { 13 }{ 6 } \right) }^{ 2 }-{ \left( \frac { 17 }{ 16 } \right) }^{ 2 } } } \)
= \(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| } +c \right] \)
= \(\frac { 1 }{ 3 } \times \frac { 1 }{ 2\times \frac { 17 }{ 6 } } log\left| \frac { x+\frac { 13 }{ 6 } -\frac { 17 }{ 6 } }{ x+\frac { 13 }{ 6 } +\frac { 17 }{ 6 } } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { x-\frac { 2 }{ 5 } }{ x+5 } \right| +c\)
= \(\frac { 1 }{ 17 } log\left| \frac { 3x-2 }{ 3(x+5) } \right| +c\)

Let I =  ഽ x³ sin (x⁴) dx
Put t = x⁴
⇒ dt = 4x³ dfx
\(\Rightarrow \frac { dt }{ 4 } ={ x }^{ 3 }dx\)
\(\therefore I=\int { sin } t.\frac { dt }{ 4 } =\frac { 1 }{ 4 } sint\quad dt\)
= \(-\frac { 1 }{ 4 } cos\quad t+c\)
= \(-\frac { 1 }{ 4 } cos\left( { x }^{ 4 } \right) +c\)      \(\left[ \because t={ x }^{ 4 } \right] \)

\(\therefore I=\int { \frac { { x }^{ 7 } }{ { x }^{ 5 }+1 } } dx\)
= \(\int { \left( { x }^{ 6 }-{ x }^{ 5 }+{ x }^{ 4 }-{ x }^{ 3 }+{ x }^{ 2 }-x+1-\frac { 1 }{ x+1 } \right) } dx\)
\(I=\frac { { x }^{ 7 } }{ 7 } -\frac { { x }^{ 6 } }{ 6 } +\frac { { x }^{ 5 } }{ 5 } -\frac { { x }^{ 4 } }{ 4 } +\frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } +x-log\left| x+1 \right| +c\)

Given f'(x) = a sin x + b cos x          ------(1)
ഽ f'(x) dx = a ഽ sin x dx + b ഽ cos x dx       [∵ f'(x) dx = f(x)]
⇒ f(x) = -a cos x + b sin x + c           -------(2)
Given f'(0) = 4
∴ (1) 4 = a sin 0 + b cos 0
⇒ 4 = a (0) + b (1)
[∵ sin 0 = 0 and cos 0 = 1]
⇒ 4 = b
Also, f(0) = 3
∴ (2)  3 = -a cos 0 + b sin 0 +c
⇒ 3 = -a (1) + b (0) +c
⇒ 3 = -a +c                                             --------(3)
And f\(\left( \frac { \pi }{ 2 } \right) \) = 5
∴ (2)  5 = -a cos \( \frac { \pi }{ 2 } \)+ b sin \( \frac { \pi }{ 2 } \)+ c
⇒ 5 = -a (0) + b (1) +c
[∵ sin \( \frac { \pi }{ 2 } \)=1 and cos \( \frac { \pi }{ 2 } \)= 0]
⇒ 5 = b + c
⇒ 5 = 4 + c      [∵ b = 4]
⇒ 5 - 4 = c
c = 1
Substituting c = 1 in (3) we get,
 3 = -a + 1
⇒ a = 1 - 3
⇒ a = -2
∴ From (2), f(x) = -(-2) cos x + 4 sin x + 1
⇒ f(x) = 2 cos x + 4 sin x +1